Mathematics Resource Guide - Instructor's Manual 2024-2025

OUR CHANGING CLIMATE Central High School - Fort Worth, TX MATHEMATICS...

OUR CHANGING CLIMATE Central High School - Fort Worth, TX MATHEMATICS An Overview of Permutations, Combinations, Algebra, and Statistics Instructor’s Manual 2 0 24 – 2 0 25 The vision of the United States Academic Decathlon® is to provide students the opportunity to excel academically through team competition. Toll Free: 866-511-USAD (8723) Direct: 712-326-9589 Fax: 712-366-3701 Email: [email protected] Website: www.usad.org This material may not be reproduced or transmitted, in whole or in part, by any means, including but not limited to photocopy, print, electronic, or internet display (public or private sites) or downloading, without prior written permission from USAD. Violators may be prosecuted. Copyright ® 2024 by United States Academic Decathlon®. All rights reserved.  Table of Contents INTRODUCTION..................3 SECTION 3: STATISTICS......... 59 3.1 Descriptive Statistics.............. 59 SECTION 1: OVERVIEW 3.1.1 Mean, Median, and Mode...........59 OF PERMUTATIONS AND 3.1.2 Range, Quartiles, and IQR.......... 65 COMBINATIONS..................4 1.1 The Multiplication Principle......... 4 3.2 Measures of Variation.............68 3.2.1 Variance........................ 68 1.2 Permutations..................... 6 3.2.2 Standard Deviation................ 72 3.2.3 Z-Score.........................74 1.3 Combinations....................11 Central High School - Fort Worth, TX 3.3 Basic Probability.................77 Section 1 Summary: Overview of 3.3.1 Independent Events................ 79 Permutations and Combinations........15 3.3.2 Dependent Events.................82 Section 1 Review Problems: Overview of 3.4 Probability Distributions...........86 Permutations and Combinations........15 3.4.1 Expected Value................... 87 3.4.2 Variance and Standard Deviation SECTION 2: ALGEBRA........... 17 of Probability Distributions..............90 2.1 Sequences and Series............. 17 2.1.1 Arithmetic and Geometric Sequences.. 19 3.5 The Binomial Distribution......... 93 2.1.2 Arithmetic and Geometric Series..... 24 3.6 The Normal Distribution..........100 2.1.3 Sigma Notation................... 31 Section 3 Summary: Statistics........ 103 2.2 Polynomials.....................33 2.2.1 Adding and Subtracting Polynomials... 34 Section 3 Review Problems: Statistics.. 106 2.2.2 Multiplying Polynomials............ 35 CONCLUSION...................111 2.3 The Binomial Expansion Theorem...38 ANSWER KEYS................. 112 2.4 Compound Interest............... 42 2.4.1 Investing and Borrowing........... 42 Section 1 Review Problems: Overview of 2.4.2 Annuities and Loans.............. 44 Permutations and Combinations—Answer Key............................. 112 2.5 Euler’s Constant................. 49 Section 2 Review Problems: Algebra— Section 2 Summary: Algebra.......... 54 Answer Key....................... 113 Section 2 Review Problems: Algebra.... 57 Section 3 Review Problems: Statistics— Answer Key....................... 117 2024–2025 Mathematics Instructor’s Manual 2  Introduction In this year’s Mathematics Resource Guide, we will study three connected areas of mathematics: permutations and combinations, algebra, and statistics. The idea that connections exist between these areas of mathematics may seem strange to you now, but connections and relationships between seemingly disconnected topics such as these are at the heart of mathematics. It is our hope that after reading through this Mathematics Resource Guide, you will be aware of and interested in these connections among different areas of mathematics. The first section of the resource guide will cover permutations and combinations—topics that are sometimes discussed in a general high school math sequence, but perhaps are only given a cursory treatment that often leaves students without a good sense of their power, flexibility, and array of applications. Combinations in particular are extremely important and are used in a wide variety of mathematical contexts, and so a comfort level with these mathematical structures will pay dividends in your future study of mathematics. Central High School - Fort Worth, TX Section 2 will focus on the topic of algebra. Algebra is a term that has seemingly become synonymous with high school mathematics, and yet the algebra most students learn in high school mathematics courses is only a small fraction of the field of algebra. Our purpose in this section is to highlight some important algebraic ideas and patterns that are often overlooked in a standard study of high school algebra. Being comfortable with arithmetic and geometric sequences and series is at least as important as knowing how to factor a quadratic, but most students spend comparatively little time looking at these sequences and series. Sigma notation for series becomes increasingly important in the study of calculus and mathematics beyond calculus, so this resource guide aims to provide you with plenty of opportunities to practice reading and manipulating sigma notation. The Binomial Expansion Theorem is another foundational topic often left out or deemphasized in traditional high school mathematics, and we will use our previous work with combinations to make sense of this important theorem. Finally, Euler’s constant, e, is often used but rarely properly understood, and so we will look at e both from a contextual and a mathematical perspective. Statistics is a branch of mathematics often only briefly covered in high school mathematics and misunderstood by much of the general public. In the final section of the Mathematics Resource Guide, we will take a more mathematical look at some of the foundations of statistics, and discuss the reasons different statistical measures, such as mean, median, variance, and standard deviation, were developed. The foundational concepts of probability distributions, the Binomial Distribution, and the Normal Distribution will also be investigated. We hope you find this year’s Mathematics Resource Guide interesting and insightful as we look at connections between these seemingly disparate areas of mathematics. Enjoy! 2024–2025 Mathematics Instructor’s Manual 3  Section 1 Overview of Permutations and Combinations In many high school mathematics sequences, permutations and combinations are often left out all together. When discussed, they are not usually given sufficient time for students to develop an appreciation and mastery of these topics. This is certainly not the fault of high school mathematics teachers—there is simply too much good content to discuss and not enough time! Yet permutations and combinations are extremely important concepts that underpin a great deal of higher mathematics. Indeed, one of the fields of current mathematical study and research is called combinatorics. Studying permutations and combinations also has great benefits for high school students. Initially these topics are fairly accessible, and some of the early problems can be approached quickly and solved easily. Yet, despite their humble beginnings, they eventually become powerful tools for solving a wide array of problems, and this flexibility makes them very useful. The study of permutations and combinations encourages and requires Central High School - Fort Worth, TX visualization, algebraic fluency, and an attention to accurate calculations—all positive mathematical traits. In last year’s Mathematics Resource Guide, some time was spent discussing permutations and combinations and relating them to probability. As we will use permutations and combinations again in this year’s Mathematics Resource Guide, we will begin with a refresher on these concepts. If you are able, you can review the portions of last year’s Mathematics Resource Guide on these topics. If not, have no fear! Everything you will need to know and understand about permutations and combinations will be discussed in this year’s resource guide. 1.1 THE MULTIPLICATION PRINCIPLE We will begin with a familiar idea from elementary school mathematics: multiplication. Multiplication is most often useful in situations of repeated addition: rather than add up 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3, which could become long and tedious, we instead use multiplication and write 3 × 9 = 27. Many times (hah!) multiplication is introduced as repeated addition, and if asked, many people will give “repeated addition” as a definition, or the definition, of multiplication. (It turns out some mathematicians may disagree with this definition, but that is another story.) Multiplication is in some ways the first important mathematical notation encountered, as it represents an operation that is a shortening of a (potentially) long process of calculation. Let’s start with some straightforward examples. EXAMPLE 1.1a: Toni is having a birthday party next week and wants to give everyone six pieces of candy in their gift bag. If Toni is inviting 15 people to the birthday party, how many pieces of candy does Toni need? SOLUTION: One way to solve this problem is to add up six 15 times: 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6. This makes sense because each of the 15 guests gets six pieces of candy. Although we might expect a first- or second-grader to solve the problem using addition, we are slightly more mathematically sophisticated, and say 15 × 6 = 90. 2024–2025 Mathematics Instructor’s Manual 4  EXAMPLE 1.1b: Toni goes to the store to purchase the candy. Unfortunately, the only candy the store has is little packages of candies with 7 pieces of candy in each package. Toni decides to give each guest six packages of this candy. How many pieces of candy does Toni purchase in order to fill the gift bags? SOLUTION: Toni was planning on getting 90 pieces of candy, since 15 × 6 = 90. But now each of the 90 packages contains 7 pieces of candy, so Toni will purchase 15 × 6 × 7 = 630 pieces of candy. EXAMPLE 1.1c: Once at the party, everyone decides they need to open their gift bag immediately after receiving it, rather than wait until they get home. It turns out the individual pieces of candy inside the packages were a bit larger than Toni thought, and they need to be eaten in two bites each. Assuming every guest does this uniformly, how many bites will it take for all fifteen guests to eat all of the candy? SOLUTION: Each of the 630 pieces of candy will take two bites to eat, so there will be a total of 15 × 6 × 7 × 2 = 1,260 bites before all the candy is eaten. Central High School - Fort Worth, TX Although this is a nice series of problems, and it certainly represents a good use of multiplication, it doesn’t seem that this solution strategy generalizes easily. Each guest has six packages of candy, and each package of candy has seven pieces, and each piece takes two bites to eat. How often do situations like this really arise? Let’s take a step back for a moment and imagine what we would do to solve this problem if we didn’t understand multiplication very well, or hadn’t learned multiplication yet. One option would be to make a list of all bites needed to eat the candy. We could make up names for each of the fifteen guests, assign the packages of candy numbers, number off the pieces of candy, and start writing: Amy, Package 1, Candy 1, Bite 1 Amy, Package 1, Candy 1, Bite 2 Amy, Package 1, Candy 2, Bite 1 Amy, Package 1, Candy 2, Bite 2 As you can imagine, this would take quite a while. After we get through all of Amy’s seven candies in Package 1, we would move to Package 2, and only after we write out all of Amy’s six packages could we even start on Bob’s candies. While writing out the entire list (with 1,260 entries!) would clearly be awful, thinking about the giant list is helpful. We can see there are 15 different possibilities for the person, 6 different possibilities for the package, 7 different possibilities for the candy, and 2 different possibilities for the bite. This mathematical structure—keeping track of the different possibilities for each entry in a list—is more applicable and can be applied to a variety of contexts. EXAMPLE 1.1d: Many license plates have three letters A–Z followed by three numbers 0–9. Assuming there are no restrictions on the letters or numbers used, how many different license plates are possible? SOLUTION: Again, we clearly don’t want to write out the entire list of possibilities, but imagining those possibilities is a good place to start. Our list would begin with the license plate AAA 000, followed by AAA 2024–2025 Mathematics Instructor’s Manual 5  001, and AAA 002. Eventually we would get to AAA 999, followed by AAB 000 (exciting!). Only after a brutally long time would we ever get to start a license plate with B, let alone get all the way down to the end of the list at Z Z Z 999. So how many license plates are possible? With 26 choices for each of the letters and 10 choices for each of the numbers, there will be 26 × 26 × 26 × 10 × 10 × 10 = 17,576,000 different possible license plates. (Good thing we didn’t try to write them all out!) Keeping these examples in mind, it seems we are ready to generalize. THE MULTIPLICATION PRINCIPLE When listing out all the possibilities for k items, the total number of entries in this list is given by n1 × n2 × n3... nk , where nk is the number of possibilities for the kth item. For example, n3 is the number of possibilities for the third item. Central High School - Fort Worth, TX In terms of the candy for Toni’s party, there were four items on the list: the person, the package, the piece of candy, and the bite. Since there were 15, 6, 7, and 2 choices, respectively, for each of these items, the answer was 15 × 6 × 7 × 2. Let’s look at one more example before we make this any more complicated. Be sure to think through why the multiplication principle makes sense; memorizing and imitating formulas doesn’t help when those formulas need to be modified to apply them to new situations. EXAMPLE 1.1e: Every morning, Jesse makes a sandwich to take to school for lunch. Jesse likes sandwiches with bread, cheese, lunchmeat, and a dressing. At the local grocery store, there are three choices for bread (white, wheat, and whole grain), four choices for cheese (cheddar, Colby jack, mozzarella, and Swiss), seven choices for lunchmeat (turkey, smoked turkey, ham, honey ham, roast beef, salami, and pastrami), and three choices for dressing (mayonnaise, mustard, and spicy mustard). How many different sandwiches can Jesse make? SOLUTION: The Multiplication Principle says the answer is 3 × 4 × 7 × 3 = 252, but let’s make sure we understand why this is the case. If our sandwich consisted of only bread, there would be three different sandwiches possible. Including cheese, each type of bread can be paired with four different cheeses, making a total of 12 possible sandwiches. Each of these 12 bread-cheese sandwiches can have one of seven lunchmeats added, bringing us to 84 sandwiches. And each of these 84 bread-cheese-lunchmeat sandwiches can have one of three dressings, for a total of 252 possible sandwiches. 1.2 PERMUTATIONS The problems we have considered thus far are nice because each category is distinct, and there is no real possibility for confusion about which category contains a given object. Usually, we would not think cheddar cheese, turkey, ham, mustard would be a legitimate sandwich in the example above. When categories potentially overlap, or when 2024–2025 Mathematics Instructor’s Manual 6 we are repeatedly picking objects from the same category, we need to be more careful. Let’s consider an example. At the beginning of math class every day, Mr. Smith selects students to write up homework problems on the board. These problems are then discussed as a class. There are 26 students in Mr. Smith’s math class, and he randomly selects a student to write up each of the first five problems. How many different ways can students be assigned to the problems? This problem is more complicated than the sandwich or candy problems because we are selecting objects from the same group each time. Mr. Smith is not selecting students from five different classes or from different groups within his class. In this way, the problem is more like the license plate problem. In that problem, we were selecting repeatedly from two different groups. Here we are repeatedly selecting objects from only one group, the class of 26 students. The first thing we have to determine is if any student can be selected more than once. In our imaginary list of all possible arrangements, is Fred, Fred, Fred, Fred, Fred an allowable selection? Or if Fred is selected to write up the first problem, is he “safe” from writing up problems #2 – 5? This is an important distinction and can drastically impact the answer to the question. Fortunately, this does not significantly alter the way we think about the problem, just the answer we get. Hopefully whether or not objects can be repeated is clear from the context and description of the problem. Sometimes, the phrases “with replacement” or “without replacement” are used to clarify whether or not an object from a category can be selected more than once. This example will be presented for you to solve in the review problems at the end of this section. Let’s now work Central High School - Fort Worth, TX through a few examples here. EXAMPLE 1.2a: How many ways can three different cards be drawn with replacement from a standard 52- card deck? SOLUTION: Since the cards are drawn with replacement, each card is put back into the deck before the next card is drawn. So, the same card could be drawn all three times, and there are 52 choices for the first card, 52 choices for the second card, and 52 choices for the third card. Therefore, there are 52 × 52 × 52 = 140,608 different ways three cards can be drawn with replacement. EXAMPLE 1.2b: How many ways can three different cards be drawn without replacement from a standard 52-card deck, assuming the order of drawing cards matters? SOLUTION: Since the cards are drawn without replacement, each card is kept when the next card is drawn. This means there are 52 choices for the first card, but only 51 choices for the second card, and 50 choices for the third card. Therefore, there are 52 × 51 × 50 = 132,600 different ways three different cards can be drawn without replacement. There are two important things to consider from this pair of problems. The first is the difference in the answers. Although there are more ways to draw three cards with replacement than without, this difference is perhaps not as large as expected. Because there are so many cards to choose from, and we are selecting so few of them, most 2024–2025 Mathematics Instructor’s Manual 7 of the entries in our imaginary list of all 140,608 ways to write out three cards with replacement will contain three different cards. Therefore, the vast majority of these entries will be included in our imaginary list of all 132,600 ways to select three different cards. It seems as we select more and more cards from the deck, this difference should become greater. The second idea that should be considered carefully is that when drawing is done without replacement, the order in which the cards are drawn makes a difference. Maybe the cards are being drawn and then placed face-up to create some form of “lineup,” but most of the time when cards are drawn from a deck, the order in which they are dealt doesn’t matter. How can this potentially more realistic situation be dealt with (hah!)? We will return to this line of questioning in a moment. For now, let’s consider the difference between selecting with replacement and without replacement (assuming order matters). EXAMPLE 1.2c: How many different ways can 52 cards be drawn from a deck with replacement (assuming the order of selection matters)? SOLUTION: Since there are 52 choices for the first card, and 52 choices for the second card, and the third card, etc., we could write 52 × 52 × 52... 52, with fifty-two 52s in this list. Rather than write out the numerical value of this calculation (which contains 90 digits), we use mathematical notation to say the answer is. Not only is Central High School - Fort Worth, TX this a nice way to write the answer, it also gives us some insight into the problem, as we can see the answer is fifty-two 52s all multiplied together. EXAMPLE 1.2d: How many different ways can 52 cards be drawn from a deck without replacement (where the order of the cards matters)? SOLUTION: Notice that eventually every card in the deck will be drawn. This means that when it is time to draw the last card, there is only one option, since every other card has already been drawn. So, there are 52 choices for the first card, but then 51 choices for the second card, and 50 choices for the third card, and so on, down to only 1 choice for the last card. Therefore, the calculation we need to perform is 52 × 51 × 50 × 49... 3 × 2 × 1. Even though this number is much smaller than and has “only” 68 digits, we still don’t want to write out this number. Furthermore, writing out this 68-digit number would not convey anything about where this number came from and would not give any insight into the problem. It seems we need some new notation, kind of like exponent notation but instead of multiplying by the same number each time, we decrease the number we are multiplying by each time, as in 52 × 51 × 50 × 49... 3 × 2 × 1. DEFINITION The symbol “!”, read aloud as “factorial,” means the product of all whole numbers starting from the number indicated down to 1. For example, 3!, read aloud as “three factorial,” means 3 × 2 × 1, so 3! = 6. 2024–2025 Mathematics Instructor’s Manual 8 So, rather than answering the question in Example 1.2d as 52 × 51 × 50 × 49... 3 × 2 × 1, we can say the answer is 52!. In addition to making the answer easier to write, this notation also helps give some insight into the solution to the problem: take all the whole numbers starting from 52 down to 1 and multiply them together. We can see now that the difference between selecting with replacement and without replacement becomes larger as we select more and more objects. Although 5252 and 52! are both very large numbers, 4.72579 × , which means the monstrous list of all 52! arrangements when repetition is not allowed takes up a miniscule 0.0000000000000000000472% of the ridiculously larger gigantic list of all arrangements if repetition is allowed. For our purposes, we will consider factorial notation as defined for positive whole numbers only. (It turns out, however, that mathematicians have discovered/invented a function that allows them to perform strange factorial calculations, like !. Weird, no?) Also, we will define 0! = 1 for reasons that may be clear shortly. Factorial notation is certainly helpful if we are selecting every member of a group, like all 52 cards in a deck. But, what if we are not selecting every object from the group? Can we still use factorial notation to help write the answer? Central High School - Fort Worth, TX EXAMPLE 1.2e: A computer program generates access codes consisting of five letters A–Z, but each letter can be used only once per access code. How many different codes are possible? SOLUTION: There are 26 choices for the first letter, but then only 25 choices for the second letter, and 24 for the third, 23 for the fourth, and 22 for the fifth. Certainly we can write 26 × 25 × 24 × 23 × 22, but this isn’t particularly satisfying and would get worse if the codes contained more letters (say, for example, 20—Bleh!). The numerical value of the answer (7,893,600) doesn’t give us any insight into the problem. How can we use factorial notation to write this number? 26! would mean 26 × 25 × 24... 3 × 2 × 1, but we don’t want all of these numbers multiplied together; we just want them down to 26 × 25 × 24 × 23 × 22. All of the numbers from 21 down to 1 need to be canceled out. To cancel out a multiplication, we use division. Therefore, we write 26 25 24 23 22 = =. Not only does this use of factorial notation make the answer easier to write, it also gives us some insight into the answer: 26 objects are in our group, and we want to select five of them without replacement (and the order of selection matters), so 21 objects are being left out. Does this idea extend to other problems? EXAMPLE 1.2f: 15-character tracking numbers are generated using letters A–Z and numbers 1–9 (0 is omitted to avoid confusion with the letter O). If each character can only be used once per tracking number, how many different tracking numbers are possible? 2024–2025 Mathematics Instructor’s Manual 9  SOLUTION: One option is to write out 35 × 34 × 33 × 32.... But, what is the last number we would need, anyway? If 15 characters are being used, that means 20 are being left out, which means there will be 21 choices for the last character. So our answer is 35 × 34 × 33... 23 × 22 × 21. Is there a better way to write this? More efficiently, we say the answer is. This type of problem and solution structure occurs frequently enough that mathematicians have given it a name: permutations. DEFINITION A permutation is an arrangement of objects from a group where no object can be used more than once and the order of selection matters. Central High School - Fort Worth, TX PERMUTATIONS FORMULA The total number of permutations of k objects from a group containing n objects is given by the formula. This formula should make good sense if we abstractly consider a problem like the tracking number problem. Given n objects, if all of the objects are arranged, this can happen in n! ways. If k of the objects are being arranged, then the first k numbers in the list n, n – 1, n – 2, etc., need to be multiplied together. This leaves n – k numbers at the end of the list to be canceled out (since k + n – k = n), ending with 1. Therefore, (n – k)! needs to be canceled out of n!, and hence. n × (n – 1) × (n – 2)... (n – k + 2) × (n – k + 1) × (n – k) × (n – k – 1)... 3 × 2 × 1 k terms n – k terms EXAMPLE 1.2g: An art director for a museum is selecting paintings to be displayed along a stretch of hallway. There is space for 5 paintings to be displayed, and the art director has 12 different paintings from which to choose. How many different ways can the art director display 5 paintings down the hallway? SOLUTION: This is an example of permutations since no painting can be chosen more than once and because the order of selection matters. (Being the first painting down the hallway is different from being the third painting down the hallway.) Therefore, there are = 95,040 different displays possible. 2024–2025 Mathematics Instructor’s Manual 10 But what if we are selecting objects and the order of selection doesn’t matter; for example, what if we are being dealt three cards out of a standard deck. Typically when we play card games, the order in which we are dealt cards doesn’t matter, only which cards are in our hand after the deal. Another classic example is selecting people to be members of a committee: the order of the people selected (usually) doesn’t matter, just who has been selected at the end. How can we count arrangements in these types of situations? 1.3 COMBINATIONS EXAMPLE 1.3a: Three cards are dealt from a standard deck of 52 cards without replacement. How many different sets of three cards are possible? (The order in which the cards are dealt does not matter.) A standard deck of cards is divided into four suits (hearts, diamonds, spades, and clubs) each of which contains 13 cards (numbers 2 through 10, jack, queen, king, and ace). SOLUTION: We already know there are = 132,600 different arrangements of three cards if the order does matter, and we have been imagining the list of all these possible arrangements. Let’s consider an entry in this list, like 2♣, 3♠, 4♦. Since we created this list of arrangements with the understanding that order mattered, these same three cards will appear in other entries in the list, just in different orders. How many entries Central High School - Fort Worth, TX will this be? Since we have three different cards, it should be 3 × 2 × 1 = 6 different times. Listing them out confirms this: 2♣, 3♠, 4♦; 2♣, 4♦, 3♠; 3♠, 2♣, 4♦; 3♠, 4♦, 2♣; 4♦, 2♣, 3♠; 4♦, 3♠, 2♣ So, this particular set of three cards was counted six times when order mattered, but should now only be counted one time since order doesn’t matter. The tricky part of this argument is convincing ourselves that this is true for every set of three cards. There should be nothing special about 2♣, 3♠, 4♦, and we could repeat the same listing of six arrangements for any set of three cards. (Convince yourself this is true.) The list of all 132,600 arrangements when order matters has therefore over-counted by a factor of 6 when order doesn’t matter, so our final answer is 132,600 ÷ 6 = 22,100 different possible sets of three cards. The key idea for solving this problem was that the list of all possible arrangements when order matters (permutations) over-counted the list of all possible arrangements when order doesn’t matter (we don’t have a name for this yet). In this case, the over-counting factor was 6 because we were selecting three objects, and those three objects could be arranged in 3 × 2 × 1 = 6 ways. But wait! 3 × 2 × 1 = 3!, so we could have written this entire calculation as. This seems like a nice formula, and each portion of the calculation seems to make sense: we have 52 objects, we are selecting 3 of them and leaving out 49 of them, and the order of selection doesn’t matter, so we need to take care of the over-counting factor. Will this strategy and formula always work on these types of problems? 2024–2025 Mathematics Instructor’s Manual 11  EXAMPLE 1.3b: A committee of four teachers needs to be selected from the 16 math teachers in a school to select a new Algebra 1 textbook. How many different committees are possible? SOLUTION: This is a mathematically similar situation to being dealt cards from a deck because no object (or in this case, person) can be selected more than once, and the order of selection doesn’t matter. If our nice formula from the previous example is going to work here, we hope the calculation is. Let’s carefully think through this problem and see if our formula turns out to be correct. If the order of selection mattered (maybe the order of selection determines the role the person will have within the committee, like chair or recorder), then this problem would be a permutation, and we would have = 43,680 different committees. But, the order of selection doesn’t matter, so our list of 43,680 permutations is too long. By what factor have we over-counted? Let’s consider one item in our list of 43,680 possible permutations: Mr. Hanks, Ms. Roberts, Mr. Jones, Ms. Wright. How many different times does this group of four teachers appear in the list of all possible permutations? Since there are four different teachers, and they will appear in every possible order, these four teachers will appear in the list of all permutations 4 × 3 × 2 × 1 = 24 times. (If you are unsure of this, write them out!) Furthermore, any set of four teachers will appear in the list of all permutations 24 times, so the Central High School - Fort Worth, TX over-counting factor is 24. Therefore, the calculation to determine the number of possible committees of four teachers is , as predicted, and there are 1,820 different possible committees. Based on the discussion of these examples, it seems we are ready to name this type of problem and generalize the calculation. DEFINITION A combination is an arrangement of objects from a group where no object can be used more than once and the order of selection does not matter. COMBINATIONS FORMULA The total number of combinations of k objects from a group containing n objects is given by the formula. 2024–2025 Mathematics Instructor’s Manual 12  NOTATION Mathematicians use the notation , read aloud as “n choose k”, for the number of possible combinations when k objects are selected from n objects. Example: There are (read aloud as “16 choose 4”) ways to select a committee of four from a group of 16 people. = , so there are 1,820 different possible committees. The similarity of the combinations formula to the permutations formula should not be surprising, as we used the permutations formula as the starting point to build the combinations formula. The important difference between a permutation and a combination is that in a combination, order does not matter. In a permutation, order does matter. This means that the list of all possible permutations over-counts the list of all possible combinations since different arrangements of objects are considered distinct in a permutation, but are all counted the same in Central High School - Fort Worth, TX a combination. Hopefully, the discussion of the two previous examples has given some informal justification to the fact that the over-counting factor is k!. Formally, if we consider a set of k objects, there are k! different ways to arrange these objects without repetition. Therefore, the set of all permutations will over-count the number of all combinations by exactly k!. Although knowing and understanding the formulas for permutations and combinations is very important, computing permutations and combinations by hand is not strictly necessary and indeed should be avoided in certain situations. (For example, should not be calculated by hand.) Almost all calculators have shortcuts or commands for combinations and permutations, and you should learn how to use these commands efficiently. Combinations are an extremely useful and flexible mathematical concept with a wide array of applications. Throughout the remainder of this Mathematics Resource Guide, we will use combinations in a variety of ways. In particular, we will focus on the use of combinations in algebra and probability. In order to give readers some idea of the utility of combinations at this time, we will conclude this section with an example using combinations in a non-routine way. EXAMPLE 1.3c: Toni has decided that giving each of the 15 guests exactly the same number of packages of candy is a bit boring and wants to mix it up a bit. Toni would like everyone to get at least two packages of candy, but the remaining packages of candy will be distributed randomly into the gift bags, so that each guest will probably end up with a different number of packages of candy. How many different ways can Toni distribute the 90 packages of candy? SOLUTION: This problem seems to be a long way from combinations. Toni could distribute the candy in a seemingly endless number of ways. For example, one person could get 62 packages of candy, and everyone else could get 2. Two people could get 32 packages of candy, and everyone else could get 2. As long as the total number of packages adds up to 90, and there are fifteen bags, this is a possible arrangement. 2024–2025 Mathematics Instructor’s Manual 13 90 packages of candy and 15 bags is quite a bit of candy to distribute. Let’s try the problem with smaller numbers and see what happens. What if Toni only has 10 packages of candy and 5 guests? Let’s say each person will receive at least one package of candy. This means there are only 5 packages of candy that can “float” from bag to bag. Let’s write out a few possibilities for these 5 packages being separated into the five different gift bags. 1, 1, 1, 1, 1 0, 2, 0, 1, 2 4, 1, 0, 0, 0 0, 0, 1, 3, 1 Although there are clearly fewer arrangements for this problem than with our original problem, the connection to combinations—or a quick way to determine the number of possibilities—still seems unclear. Let’s say Toni has the five different gift bags in a row and has the five packages of candy that are allowed to “float” in hand. Toni will put some number of packages in the first gift bag (possibly 0), move to the second gift bag, put some number of packages in the second gift bag (again, possibly 0), and so on down the line. If p represents placing a package of candy in the gift bag, and n represents moving to the next gift bag, then each distribution of the 5 packages can be represented by a string of 5 p’s and 4 n’s. For example: Central High School - Fort Worth, TX 1, 1, 1, 1, 1 can be represented by pnpnpnpnp 0, 2, 0, 1, 2 can be represented by nppnnpnpp 4, 1, 0, 0, 0 can be represented by ppppnpnnn 0, 0, 1, 3, 1 can be represented by nnpnpppnp Since there is a one-to-one correspondence between the arrangements of p’s and n’s and the distributions of the packages of candy, there is the same number of arrangements as distributions. With 9 spots to be filled, five of them need to be selected for p. (The remaining spots will automatically be filled by n’s.) No spot can be picked more than once, and the order of selection doesn’t matter. Ah ha! A combination! There are therefore = 126 different arrangements of 5 p’s and 4 n’s, and this means there are 126 ways to fill 5 different gift bags with 10 packages of candy, assuming each bag must have at least one package. Can this thought process be extended to our original problem? With 15 gift bags, if each gift bag must have at least two packages of candy, 60 packages of candy are allowed to “float” from bag to bag. If we imagine the 15 gift bags in a row, Toni will put some number of packages of candy in the first gift bag (possibly 0), move to the second gift bag, put some number of packages of candy in the second bag (possibly 0), move to the third bag, and so on. At the end of the process, Toni will have placed 60 packages of candy and moved to the next gift bag 14 times. We can then imagine a tremendously long string of p’s and n’s representing each possible placement of the candy as Toni moves from bag to bag. Each string will contain 74 total characters: 60 p’s and 14 n’s. Therefore, we must select 60 of the 74 places for the p’s, and these places are selected without replacement, and the order of selection does not matter. Again we have a combination! Therefore, there are = 87,178,291,200 different ways Toni can distribute the 90 packages of candy to the 15 gift bags, assuming each gift bag contains at least two packages of candy. Whew! It is a good thing we didn’t try to list them all out! 2024–2025 Mathematics Instructor’s Manual 14 This example illustrates the common “lineup” representation and use of combinations to count the total number of possible arrangements in said lineup. As we will see in other sections of this resource guide, combinations are a useful mathematical tool with a wide array of applications. At this point, we hope you have some sense of when a combination might be useful and knowledge of how to properly calculate combinations. SECTION 1 SUMMARY: OVERVIEW OF PERMUTATIONS AND COMBINATIONS The Multiplication Principle: When listing out all the possibilities for k items, the total number of entries in this list is given by n1 × n2 × n3 × × × nk , where nk is the number of possibilities for the kth item. (For example, n3 is the number of possibilities for the third item.) Factorial Notation: The symbol “!”, read aloud as “factorial,” means the product of all whole numbers starting from the number indicated down to 1. For our purposes, factorial is only defined for positive whole numbers and 0. By definition, 0! = 1. A permutation is an arrangement of objects from a group where no object can be used more than once, and the order of selection matters. Permutations Formula: The total number of permutations of k objects from a group containing n Central High School - Fort Worth, TX objects is given by the formula. A combination is an arrangement of objects from a group where no object can be used more than once, and the order of selection does not matter. Combinations Formula: The total number of combinations of k objects from a group containing n objects is given by the formula. Combinations Notation: Mathematicians use the notation , read aloud as “n choose k,” for the number of possible combinations when k objects are selected from n objects. Therefore, =. SECTION 1 REVIEW PROBLEMS: OVERVIEW OF PERMUTATIONS AND COMBINATIONS For each of the problems below, identify whether the problem should be solved using the Multiplication Principle, permutations, combinations, or some mixture of these three methods. Explain your reasoning, and then solve the problem. 1.  t the beginning of math class every day, Mr. Smith selects students to write up homework problems on A the board. These problems can be discussed as a class. There are 26 students in Mr. Smith’s math class, and he randomly selects with replacement a student to write up each of the first five problems. How many different ways can students be assigned to the problems? 2.  r. Smith’s students eventually complain that it isn’t fair that a student can be selected more than once M and can be selected to write up all five problems. Mr. Smith agrees that he will now select the five students each day without replacement. How many different ways can students be assigned to the problems? 2024–2025 Mathematics Instructor’s Manual 15 3.  r. Smith’s students now comment that a student may be selected for a problem they are not able to M do correctly and propose an alternate selection method: a group of five students is selected at random, and then these five students decide amongst themselves who will write up each problem. Mr. Smith eventually agrees. How many different ways can Mr. Smith select a group of five students? 4. A regular polygon is a polygon with equal angle measures and equal side lengths. A diagonal of a polygon connects two non-adjacent vertices. How many diagonals are there in a regular heptadecagon (17-sided polygon)? 5. In a computer game for children, a picture of scenery (like a mountainside) is divided into 8 regions. There are 6 different choices of color, and any region can be painted any color. How many different ways is it possible to color a given picture? 6. Ernest is traveling to New York City and has created a list of ten different possible sightseeing activities in which he is interested. He will be in New York City for three days, but will only have time for two different activities each day. How many different sightseeing plans can Ernest create? (Assume each day Central High School - Fort Worth, TX is treated separately, and clearly Ernest will not want to complete each activity more than once.) 7. The programming director of a local television station is setting the schedule for the upcoming Saturday. On Saturdays, this station shows a series of five movies taking up the morning and afternoon programming hours. If the programming director has 45 movies from which to choose, how many different movie schedules are possible for this Saturday? 8. What is the coefficient of in ? 2024–2025 Mathematics Instructor’s Manual 16  Section 2 Algebra Unlike permutations and combinations, algebra is a topic that usually receives a great deal of discussion in a typical high school mathematics sequence. As with all branches of mathematics, however, many ideas remain beyond what most students encounter during their high school years. In this section of the resource guide, we hope to broaden the scope of traditional high school algebra and show how some of the different mathematical ideas in and around algebra are related to each other. We assume the reader has some familiarity and fluency with topics traditionally covered in high school algebra courses, such as solving linear equations, basic factoring, solving quadratics, and the quadratic formula. Rather than rehash topics such as these, which are typically discussed in traditional high school math classes, we will focus on some topics that may not receive as much attention but are still critically important in higher mathematics: sequences and series, polynomials, the Binomial Expansion Theorem, compound interest, and Euler’s constant. Central High School - Fort Worth, TX 2.1 SEQUENCES AND SERIES Mathematics, in part, is the study of patterns and attempts to recognize, extend, and classify these patterns. They may occur geometrically, graphically, or algebraically, but many of the patterns in which mathematicians are interested are numerical ones. From an early stage in our mathematical studies, we are familiar with lists of numbers and attempts to extend these lists. EXAMPLE 2.1a: What are the next three numbers in the following list: 2, 4, 6, 8 …? SOLUTION: As this seems to be a list of even numbers, the next three numbers are 10, 12, 14. Mathematicians call a list of numbers like this a sequence. DEFINITION A sequence is a list of objects presented in a particular order. The objects in the sequence are called the terms of the sequence. Most sequences are made up of numbers, as in Example 2.1a, and most sequences we will encounter in this resource guide will be numerical sequences, but strictly speaking, the terms of a sequence do not need to be numbers. Mathematicians often study sequences of different sorts of objects. 2024–2025 Mathematics Instructor’s Manual 17 Consider, for example, what object comes next in the following sequence: Or, for another example, what object comes next in the following sequence: Monday, Tuesday, Wednesday, Thursday…? All of the sequences we will consider are “nice” in that the sequence can be extended in a logical manner. Mathematicians do consider lists that do not have a pattern or formula to be sequences (a sequence of random numbers, for example), but we will restrict ourselves to nice sequences that we can extend logically and predict. Even within the realm of numerical sequences that can be extended, there are problems assuming that sequences are what they first appear to be. Any numerical sequence can, in theory, be extended in an infinite number of different ways that may or may not agree with the perceived rule or pattern for the sequence. For example, Central High School - Fort Worth, TX we said earlier that 2, 4, 6, 8, … appears to be a list of even numbers, and so extended it with 10, 12, and 14. However, this sequence does not have to be a list of even numbers, and there are many different ways to extend this sequence. The sequence 2, 4, 6, 8, 10, 58, 252, 734… is a valid mathematical sequence, although it is very different from how we expect a sequence beginning with 2, 4, 6, 8 to continue. In this resource guide, if a sequence seems to follow a simple pattern, we will generally assume it continues this pattern, but strictly speaking this is not necessarily the case. In order to communicate effectively about sequences, mathematicians have developed some common notation. NOTATION The position of a term in a sequence is called the index of the term. The terms of a sequence are denoted by a variable (usually a or x) and an index, with the index written as a subscript. Unless otherwise noted, the index begins at 1 and consists of counting numbers. For example, a generic sequence will commonly be written as x1, x2, x3… or a1, a2, a3... For the sequence of even numbers given earlier, x1 = 2, x2 = 4, x3 = 6, x4 = 8, etc. Often when we write a sequence in this form, it becomes apparent that there is a relationship between the index and the term of the sequence. In the sequence of even numbers, the term is always twice the index. Mathematicians will use another variable, usually i or k, to represent the index, and xi or xk will be used to represent the ith or kth term in the sequence. When this is done, a formula can be written to describe the relationship between the index and the term. We will use curly brackets {} to denote that an equation represents a sequence of terms, rather than an equation we might try to solve. For example, the sequence of even numbers can be written as {xi = 2i}. 2024–2025 Mathematics Instructor’s Manual 18 The curly brackets tell us we are talking about a sequence, and the index of this sequence is represented by i. With no other indication about what values i takes, we assume i begins at 1 and counts upward. When i = 1, x1 = 2 × 1, so x1 = 2. When i = 2, x2 = 2 × 2, so x2 = 4. When i = 3, x3 = 2 × 3, so x3 = 6. This pattern continues, so this sequence is the sequence of even numbers. EXAMPLE 2.1b: What sequence is generated by {xi = i 2}? SOLUTION: When it is not clear what a sequence is, writing out several terms is always a good place to start. When i = 1, x1 = , so x1 = 1. When i = 2, x2 = , so x2 = 4. When i = 3, x3 = , so x3 = 9. When i = 4, x4 = , so x4 = 16. The first four terms of the sequence are 1, 4, 9, 16, and the fifth term will be 25 (verify!), so this is the sequence of square numbers. EXAMPLE 2.1c: What formula will generate the sequence 3, 7, 11, 15, 19, …? SOLUTION: For this sequence, x1 = 3, x2 = 7, x3 = 11, and x4 = 15. It seems the index is multiplied by 4 and then one is subtracted to give the term. So, the formula that generates this sequence is. Central High School - Fort Worth, TX Although all of the sequences we have considered so far go on forever, or are infinite, many sequences in which we are interested do not go on forever, or are finite. A finite sequence only has a certain number of terms, so the index will have a starting value and ending value. We denote these limits on the index by using a superscript and subscript after the second curly bracket. The lower number represents the starting value for the index (sometimes accompanied by “i = ”), and the upper number represents the ending value for the index. EXAMPLE 2.1d: Write out the terms of the finite sequence. SOLUTION: Simplifying the square roots as we go, this sequence is 1, , , 2, , , , , 3. Some sequences have no particular pattern, or clear relationship of one term to the next, and a generating formula must be “guessed” or intuited somehow. Two particular types of sequences, however, have nice relationships from one term to the next, and therefore have nice generating formulas. These types of sequences occur frequently and have nice mathematical characteristics, which suggests they merit their own terminology and careful study. 2.1.1 Arithmetic and Geometric Sequences Sequences are particularly nice if there is a relationship that is easy to identify between one term and the next. Let’s take a look at an example: What is the apparent relationship between the terms in the sequence 3, 7, 11, 15, 19…? As mentioned previously, just because a sequence appears to have a pattern or relationship does not necessarily mean that pattern holds in the sequence. If we assume this sequence behaves as it seems to, the relationship is easy to identify: to get from one term to the next, we add 4. We have been discussing formulas that generate the terms of a sequence, like {xi = 4i − 1} or. But there is another way to represent a sequence as a formula, and that is to describe how the sequence changes from one 2024–2025 Mathematics Instructor’s Manual 19 term to the next. In this case, the sequence is adding 4 to go from one term to the next. Mathematically, we write xi = xi−1 + 4. Since represents the term in the place in the sequence, xi−1 represents the term in the (i − 1)th place in the sequence, or the previous term. This formula alone, xi = xi−1 + 4, is not enough to describe the sequence since it only describes how to get from one term to the next. This relationship could describe any of a whole group of different sequences that add 4 each time, but begin at different values. Therefore, to properly describe a sequence, this type of formula also needs a declaration of the starting value. In the example above, we would declare. This type of formula for a sequence is called a recursive formula. DEFINITION A recursive formula for a sequence is a formula that declares the starting value (or values) for the sequence and how the subsequent terms are made from the previous term (or terms). EXAMPLE 2.1e: Write a recursive formula for the sequence 3, 7, 11, 15, 19…. Central High School - Fort Worth, TX SOLUTION: The recursive formula is x1 = 3; xi−1 + 4. EXAMPLE 2.1f: Write a recursive formula for the sequence 3, 6, 12, 24, 48 …. SOLUTION: The first term of this sequence is also 3; therefore,. What is happening as we move from term to term in this sequence? It appears as though each term is twice the previous term, so xi = 2 × xi−1. Therefore, the recursive formula is ;. Recursive formulas are sometimes easier to write than formulas that generate a sequence directly, like {xi = 4i − 1}. These types of formulas are called direct formulas. However, to find the 100th term of a sequence using a recursive formula requires writing out the first 99 terms of this sequence, a potentially tedious task. To find the 100th term of a sequence using a direct formula, we substitute i = 100 into the formula. The advent of computing technology has somewhat lessened this drawback of recursive formulas, as computers can now compute thousands of terms using a recursive formula fairly quickly. Indeed, a great deal of computer programming uses recursive formulas, and some sequences are much easier to write using recursive formulas than direct formulas. EXAMPLE 2.1g: What are the terms of the sequence given by the following recursive formula: a1 = 1, a2 = 1; ai = ai−1 + ai−2? SOLUTION: Since they were declared, the first two terms of the sequence are and. What do we make of the recursive declaration in this formula? We already know and , so letting i = 3 should allow 2024–2025 Mathematics Instructor’s Manual 20  us to find a3. By substitution, a3 = a3−1 + a3−2, so a3 = a2 + a1. Therefore, a3 = 2. When i = 4, a4 + a2, so a4 = 3. When i = 5, a5 = a4 + a3, so a5 = 5. Thus far our sequence is 1, 1, 2, 3, 5, and we are able to understand what the recursive rule is doing: each term is found by adding the two previous terms together. Writing out more terms of the sequence gives us 1, 1, 2, 3, 5, 8, 13, 21, 34, …. This famous sequence is called the Fibonacci sequence. Although somewhat strange at first, the recursive formula for the Fibonacci sequence is fairly straightforward; the direct formula, on the other hand, is difficult to determine and extremely complicated! Let’s return to less complicated sequences. A sequence where a fixed amount is added to move from one term to the next (like 3, 7, 11, 15, 19, …) is called an arithmetic sequence. DEFINITION An arithmetic sequence is a sequence with a constant difference between consecutive terms. Although we generally think of this constant difference as a positive number, this is not strictly necessary, and sequences with a constant negative difference are also arithmetic. Sequences with a common difference of 0 are Central High School - Fort Worth, TX also technically considered arithmetic although these sequences aren’t very interesting. (Why not?) EXAMPLE 2.1h: What is the constant difference between the terms of the arithmetic sequence 13, 6, 1, 8, …? SOLUTION: The constant difference is −7. NOTATION FOR ARITHMETIC SEQUENCES The constant difference for an arithmetic sequence is usually represented by the variable d. The first term of an arithmetic sequence is usually represented by the variable a. This notation allows us to develop recursive and direct formulas for arithmetic sequences. We began this discussion with recursive formulas precisely because the recursive formula for an arithmetic sequence is so nice. The first term is a, so = a. To move from one term to the next, we add d, so. RECURSIVE FORMULA FOR AN ARITHMETIC SEQUENCE The recursive formula for an arithmetic sequence with an initial value of a and a constant difference of d is = a;. 2024–2025 Mathematics Instructor’s Manual 21 Using the recursive formula to write out the terms of the generic arithmetic sequence gives us a, a + d, a + 2d, a + 3d, a + 4d, …. A direct formula relies on the relationship between the index and the term, so we consider x1 = a, x2 = a + d, x3 = a + 2d, x4 = a + 3d, and x5 = a + 4d. It seems each term is a plus d times one less than the index. This makes sense if we think about what happens as we move along in the sequence. To move from the first term to the fifth term, we will add the difference, d, four times to the starting value, a. Therefore, the fifth term will be a + 4d. Generalizing this argument, to move from the first term to the kth term, we will add the difference, d, k −1 times to the starting value, a. Therefore, the kth term will be a + (k −1) × d. DIRECT FORMULA FOR AN ARITHMETIC SEQUENCE The direct formula for an arithmetic sequence with an initial value of a and a constant difference of d is = a + (k −1) × d. It makes sense that the direct formula for an arithmetic sequence would involve a multiplication by the constant difference, as this difference is added repeatedly to move from term to term, and repeated addition can be expressed as multiplication. Central High School - Fort Worth, TX EXAMPLE 2.1i: What is the 201st term in the arithmetic sequence 41, 38, 35, 32, …? SOLUTION: The first term of the sequence is 41, and the common difference is −3. To move from the 1st term to the 201st term in the sequence, the common difference will be added 200 times, so a total of 600 will be subtracted from 41. Therefore, the 201st term is −559. Note that using the direct formula is precisely the same logic, just written more formally: = 41 + (201 − 1) × (−3), so = −559. EXAMPLE 2.1j: What is the first term of the arithmetic sequence with = 136 and = 205? SOLUTION: The first thing we need to do is to determine the constant difference. From the 54th term to the 77th term, the difference is added 23 times, and the distance between = 136 and = 205 is 69. Therefore, the constant difference is 3. From the first term to the 54th term, the difference was added 53 times. This means the sequence increased 159 from the first term to the 54th term. Therefore, the first term is −23. The thought process in this example is a bit of an informal argument. We can use the direct formula several times in a more mathematically formal way to solve the same problem. Substituting known values into = a + (k − 1) × d gives us two equations: 136 = a + 53 × d (when k = 54) and 205 = a + 76 × d (when k = 77). This is then a system of two equations with two unknowns that can be solved using substitution or elimination. For example, subtracting the two equations gives us 69 = 23d, which implies d = 3. Substituting into the first equation then yields 136 = a + 53 × 3, which solves as a = −23. The other (mathematically) simple thing to do as we move from one term to the next in a sequence is multiply, as in the sequence 3, 6, 12, 24, 48, … given in Example 2.1f. These types of sequences are called geometric sequences. 2024–2025 Mathematics Instructor’s Manual 22  DEFINITION A geometric sequence is a sequence with a constant ratio between consecutive terms. Although we usually consider multiplication and division as separate operations, we know that division problems can be stated as multiplication problems and vice versa. Therefore, geometric sequences are always phrased as a multiplication from term to term, although sometimes this ratio is a fraction between 0 and 1. EXAMPLE 2.1k: What is the ratio for the geometric sequence 36, 12, 4, , …? SOLUTION: To move from term to term in this sequence, we divide by 3, so the ratio is. Ratios for geometric sequences may also be negative numbers or 1. (What would these sequences look like?). The only number that is not allowed as a ratio for a geometric sequence is 0. (Why?) Central High School - Fort Worth, TX NOTATION FOR GEOMETRIC SEQUENCES The constant ratio for a geometric sequence is usually represented by the variable r. The first term for a geometric sequence is usually represented by the variable a. As with arithmetic sequences, this notation allows for the development of recursive and direct formulas for geometric sequences. Since the first term of a geometric sequence is a, x1 = a. Since a geometric sequence multiplies by the ratio from term to term,. RECURSIVE FORMULA FOR A GEOMETRIC SEQUENCE The recursive formula for a geometric sequence with an initial value of a and a ratio of r is ;. We can use this recursive formula to write out several terms and find a direct formula. The terms of the generic geometric sequence are a, a × r, a × , a × , a × …. A direct formula finds a relationship between the index and the term, so we consider = a, = a × r, = a × , = a × , and = a ×. It seems that each term is a times r raised to a power that is one less than the index. Therefore, the kth term will be a times r raised to the k − 1 power, and = a ×. DIRECT FORMULA FOR A GEOMETRIC SEQUENCE The direct formula for a geometric sequence with an initial value of a and a ratio of r is =a×. 2024–2025 Mathematics Instructor’s Manual 23 It makes sense that this formula contains a power of r, as a geometric sequence involves repeated multiplication by r to move from one term to the next. Since repeated multiplication can be represented as an exponent, we anticipate the direct formula would have a power of r. This power should be k − 1, since we need one power of r for every term except for the first term. EXAMPLE 2.1l: What is the 20th term in the geometric sequence , , , …? SOLUTION: As the first term is and the ratio is 2, the 20th term is given by = × 220−1, so the 20th term 131,072 is 5. EXAMPLE 2.1m: What is the 4th term in a geometric sequence if the 2nd term is 2 and the 6th term is 8? SOLUTION: Since a geometric sequence multiplies by a constant ratio to move from term to term, this multiplication occurs 4 times to move from the 2nd term to the 6th term. This means that = 4 , since the 2nd term times 4 is the 6th term. So, = 2 and r =. We will multiply by the constant ratio twice to move Central High School - Fort Worth, TX from the 2nd term to the 4th term, so the 2nd term will be multiplied by = 2, and therefore the 4th term in this sequence is 4. (The first six terms of the sequence are , 2, 2 , 4, 4 , 8.) This is again a bit of an informal argument, and it can be made more formal by using the direct formula for a geometric sequence. As = 2, 2 = a × r, and as = 8, 8 = a ×. Dividing these two equations gives us = 4, and so r =. We are looking for , which is a ×. Rather than solving for a, we use 2 = a × r to build a × by multiplying both sides by. This gives us 2 × =a×. As r = , = 2, so 4 = a × , and = 4. Arithmetic and geometric sequences are mathematically important due to their simplicity and ease of use. They also behave nicely when summing the terms of a sequence, which we will investigate shortly. Arithmetic and geometric sequences are important in modeling physical phenomena and in other applications. Although most models are neither precisely arithmetic (linear) nor geometric (exponential), many growth or decay situations can be modeled extremely well using these simple equations. Some of the models encountered in typical high school mathematics are perhaps a bit simplistic, but these models are extremely powerful and far reaching. For example, the idea of the recursive step for the geometric sequence, , forms a large part of the study of differential equations in higher mathematics. 2.1.2 Arithmetic and Geometric Series In addition to being mathematically simple and having nice recursive and direct formulas, arithmetic and geometric sequences have another important mathematical property: it is possible to find a nice formula for the sum of an arithmetic (or geometric) sequence. This sum is called a series. DEFINITION A series is the sum of the terms in a sequence. 2024–2025 Mathematics Instructor’s Manual 24 Although it is always possible to find a finite series by brute computational force (determining all the terms in the sequence and then summing), this is not a very mathematical way to proceed, and mathematicians try to avoid doing such things whenever possible. The fact that there is a nice way to find the sum of arithmetic and geometric series further highlights their mathematical importance. We will begin by looking at an extremely simple, but very important, arithmetic series. EXAMPLE 2.1n: What is 1 + 2 + 3 + … + 97 + 98 + 99 + 100? SOLUTION: The first thing we notice is that this series, like most we will study, is finite; that is, it has a certain number of terms (in this case, 100). Many sequences we consider are infinite (go on forever), but most infinite series diverge, or add up to infinity. For example, any infinite arithmetic series adds up to infinity (like 3 + 7 + 11 + 15 + …) or negative infinity (like 41 + 38 + 35 + 32 + …). However, here we have a finite number of terms, so these 100 numbers certainly add up to some number. We would like a clever way to add up all 100 of these numbers, rather than just 1 + 2 = 3, 3 + 3 = 6, 6 + 4 = 10, etc. This method is rather boring, and with 99 calculations to perform, we are likely to make a mistake. Rather than adding 1 + 2, why not add 1 + 100? 2 can then be paired with 99, 3 with 98, and so on. All of these pairs have the same sum, 101, and there are precisely 50 pairs (since there are 100 numbers and we are pairing them up). Therefore, this sum is equal to 101 × 50 = 5,050. Central High School - Fort Worth, TX According to mathematical legend, the great mathematician Carl Gauss (1777–1855) was in his second-grade class when the teacher gave this problem in the hopes of having thirty minutes of peace and quiet as all his pupils painstakingly carried out all ninety-nine sums. Gauss thought about the problem for a bit and then wrote down nothing but the correct answer, having paired the numbers and multiplied in his head. Whether or not this is true, it exemplifies the idea of work smarter, not harder. Will this pairing idea work on any arithmetic series? If so, it seems we have discovered a powerful tool for finding arithmetic series. EXAMPLE 2.1o: Find the arithmetic series 3 + 7 + 11 + 15 + … + 399 + 403 + 407. SOLUTION: The pairing idea certainly starts off nicely: 3 + 407 = 410, 7 + 403 = 410, 11 + 399 = 410, and so on. But how many pairs of 410 are created by the terms in this series? We need to determine how many terms are in the sequence 3, 7, 11, 15 … 399, 403, 407. The sequence has a common difference of 4, and a total of 404 is added from the first term to get to the last term. This means 4 is added 101 times to get from the first term to the last term, which means there are 102 terms in the sequence (as 4 was not added to get the first term, 3). Therefore, there are 51 pairs of 410 in this series, and the series totals to 410 × 51 = 20,910. Does this series have to be arithmetic for this idea to work? Does this work on other types of series? 2024–2025 Mathematics Instructor’s Manual 25  EXAMPLE 2.1p: Determine the sum 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64. SOLUTION: Although we could just add up all these terms, we are interested in whether or not the pairing strategy works for non-arithmetic series, so the numerical answer here is not the point of the problem. This series is certainly non-arithmetic, since the difference in the first two terms is 3, but the difference between the 2nd and 3rd term is 5, and then 7, etc. Will the pairing strategy work? 1 + 64 = 65, but 4 + 49 = 53. Uh oh. 9 + 36 = 45, and 16 + 25 = 41. None of these pairs have the same sum, so our pairing strategy will not work on this series. Why does the pairing strategy work on arithmetic series? Since the difference between consecutive terms is always a constant, d, the pair sum increases by d as the index increases by one (we move one term up the sequence). Meanwhile, the pair sum decreases by d as the index decreases by one (we move one term down in the sequence). Therefore, the paired sum remains constant throughout, and our pairing strategy works. Can we write a formula that describes this pairing strategy for arithmetic series? If we have an even number of terms, the formula should work out nicely. With an even number of terms, we know each term in our series has a pair, and it is easy to determine the number of paired sums. Therefore, we will state our formula at first only for arithmetic series with an even number of terms. Central High School - Fort Worth, TX STRATEGY AND FORMULA FOR AN ARITHMETIC SERIES WITH AN EVEN NUMBER OF TERMS To find an arithmetic series with an even number of terms, we create paired sums equal to the first term plus the last term. If there are k terms, there will be pairs, so the arithmetic series will be ×. As = a and = a + (k − 1) × d, this formula can also be written as × , although this lacks some of the mathematical aesthetic of the first formula, as it involves the same number of variables and masks the idea that generated the formula. What if the arithmetic series has an odd number of terms? In such cases, our pairing strategy will work for the most part, but one term in the series will be left out and not have a pair. How can we deal with this problem? It seems there are three main ways to deal with this problem. One idea is to add an extra term to the series (giving it an even number of terms) and then subtract the additional term off to return to the original series. Another idea is to add up all of the terms except for the last term, so that we have an even number of terms, and then add on the term we left off. The last way is to pair up the terms as usual and then try to figure out the middle term that is left over and has no pair. We encourage you to try the second and third strategy; we will focus on the first. Let’s try our idea with a specific arithmetic sequence to see how it works before we try it in general. 2024–2025 Mathematics Instructor’s Manual 26  EXAMPLE 2.1q: What is the arithmetic series 41 + 38 + 35 + … + (−1) + (−4) + (−7)? SOLUTION: The constant difference for this series is −3, and there are 17 terms. (Is this correct? How do we know this?) Although our pairing strategy will work, it would be much nicer if there were an additional term in this series, so there would be 18 terms and 9 pairs. Therefore, we will add –10 to the end of the series, find the sum, and then add 10 to remove this extra term we included. 41 + 38 + 35 + … + (−1) + (−4) + (−7) + (−10) has a paired sum of 31, and there will be 9 of these pairs, so the series is 31 × 9 = 279. But this includes the −10 we added, so our original arithmetic series is 289. What if we hadn’t worried about whether this series had an even or odd number of terms, and just used × ? 41 + (−7) = 34? There would be 8.5 “pairs” since 17 ÷ 2 = 8.5. And 34 × 8.5 = 289! Amazing! The formula we wrote for an even number of terms seems to work for an odd number of terms as well! Is this just coincidence? Let’s try our strategy of including an extra term in general and see what happens. EXAMPLE 2.1r: What is the arithmetic series with an odd number of terms? SOLUTION: Since we’re doing this problem in general, we’ll start with = a, = a + d, and so on up to Central High School - Fort Worth, TX = a + (k − 1) × d, where k is odd. Adding one additional term at the end of this sequence will give us an even number of terms, so we’ll include xk + 1 = a + k × d. Now we consider the arithmetic series a + (a + d) + (a + 2d) + … + [a + (k − 1) × d] + (a + k × d). This series has k + 1 terms, which is even as k is odd, so this series will equal (x1 + xk + 1) × , or (a + a + k × d) ×. However, this is not the series in which we are interested, and we need to remove the term we added on. Therefore, our original series will equal (a + a + k × d) × − (a + k × d). At this point some algebraic manipulation seems necessary: − a − k × d (we write the first term as a single fraction and distribute the negative across the second term) − a − k × d (expanding the numerator of the fraction) − − (creating a common denominator, so the terms can be combined into one fraction) (subtracting the fractions) (factoring out the common factor of k from each term) (2a + kd − d) × (rewriting multiplication) [2a + (k − 1) × d] × (factoring out d from the second and third terms inside parentheses) [a + a + (k − 1) × d] × (splitting apart 2a into a + a) [ ]× (substituting = a and = a + (k − 1) × d) But this is exactly the same formula we had when the series had an even number of terms! Amazing! So, the same formula works whether our arithmetic series has an even number of terms or an odd number of terms. 2024–2025 Mathematics Instructor’s Manual 27  FORMULA FOR AN ARITHMETIC SERIES An arithmetic series with k terms is equal to ×. EXAMPLE 2.1s: An arithmetic series equals 624. The first term of this series is 3, and the second term is 5. What is the last term in this series? SOLUTION: We know that = 3 and d = 2. The fact that the series equals 624 means that × = 624, but we only know and are trying to find. As = a + (k − 1) × d in general, for this problem = 3 + 2 × (k − 1), or = 1 + 2k. Substituting in now yields (3 + 1 + 2k ) × = 624 , or (4 + 2k) × k = 1,248. This is a quadratic in terms of k, so rearranging gives us + 4k − 1248 = 0 or + 2k − 624 = 0. A quick check of the factors of 624 reveals 26 and 24, so (k + 26) × (k − 24) = 0 , and k = −26 or k = 24. Certainly our arithmetic series should have a positive number of terms, so there are 24 terms in our series. This means the last term = 1 + 2k is 49. Central High School - Fort Worth, TX What about geometric series? As we just saw, the pairing trick only works for arithmetic series. Can something be done to make geometric series easy to find? EXAMPLE 2.1t: Find the geometric series 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384. SOLUTION: Although this series does not have very many terms, and we could just add them all up, this would not lend itself to a general method. Let’s try something else. To find an arithmetic series with an odd number of terms, we added on the next term, which changed the series into something nicer. Then we went back from that series to the original series in which we were interested. Is adding a term to this geometric series helpful? We are interested in 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384, so let’s call this number S. Consider 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 + 768. This is clearly S + 768, so: 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 + 768 = S + 768. Subtracting 3 from both sides yields: 6 + 12 + 24 + 48 + 96 + 192 + 384 + 768 = S + 768 3. Now every term on the left-hand side shares a factor of 2, so factoring this out gives us: 2(3 + 6 + 12 + 24 + 48 + 96 + 192 + 384) = S + 768 – 3. But wait! 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 = S, so: 2S = S + 768 – 3. This equation can easily be solved for S, and so this geometric series is 765. Amazing! But will this trick always work? 2024–2025 Mathematics Instructor’s Manual 28  EXAMPLE 2.1u: Find the geometric series 36 + 12 + 4 + + + + +. SOLUTION: Let 36 + 12 + 4 + + + + + = S. The ratio for this geometric series is , so the next term would be. Let’s add this term to our series and then try to rewrite both sides including S. 36 + 12 + 4 + + + + + + =S+ 12 + 4 + + + + + + =S+ − 36 Now the tricky part. We need to rewrite the left-hand side as the original series. In order to turn 12 into 36 and 4 into 12, we need to factor out and write this as: 36 + 12 + 4 + + + + + =S+ − 36 S=S+ − 36 Central High School - Fort Worth, TX 26,240 13,120 Solving for S gives us: 729 = S, so S = 243. Let’s try this strategy in general on a generic geometric series and see if we can develop a formula. EXAMPLE 2.1v: What is a geometric series with k terms and a ratio of r? SOLUTION: Consider the generic geometric series a + a × r + a +…+ a × +a×. Let this equal S. The next term will be a × , so: a + a × r + a × +…+ a × +a× +a× =S+a×. Moving the a to the other side of the equation gives us: a×r+a× +…+ a × +a× +a× =S+a× − a. Every term on the left-hand side now has a factor of r, so factoring this out yields: r × (a + a × r +…+ a × +a× +a× )=S+a× −a But a + a × r + a × +…+ a × +a× is S, so: r×S=S+a× −a Now we solve for S: r×S−S=a× −a S × (r − 1) = a × −a S= 2024–2025 Mathematics Instructor’s Manual 29  FORMULA FOR A FINITE GEOMETRIC SERIES A geometric series with first term a, ratio r, and with k terms is equal to S =. Alternatively, we can write this formula by referencing the terms in the sequence: S = , where is the last term in the series and is the next term in the sequence (but is not included in the series). This formula does not hold if the ratio is equal to 1 because the method used does not work. (Why not? What is the sum if r = 1?) 1 EXAMPLE 2.1w: What is the value of the geometric series 1 + + +...+ + 1,024 ? SOLUTION: Although we could determine the number of terms in this series and use S = , since 1 1 we know the last term in this series, we will use S =. The ratio is , and as = 1,024 , = 2,048. Central High School - Fort Worth, TX 2,047 Therefore, the series equals , = 1,024. This example brings up an interesting question: what happens if our geometric series has an infinite number of terms? Clearly if an arithmetic series has an infinite number of terms, the series will equal positive infinity (if d 0) or negative infinity (if d 0). If the absolute value of a geometric series increases (if 1), then the absolute value of the series will equal positive infinity. But, what happens if the terms in a geometric series get closer to 0 (if 0), as in this example? EXAMPLE 2.1x: What is the value of the infinite geometric series 1 + + + +...? SOLUTION: Since this is an infinite geometric series with a ratio of , the terms of the sequence are approaching 0. This means the “last term” in this series is 0. (Technically, there is no last term since the series goes on forever, but the terms get close enough to 0 that we can’t tell the difference. This will seem a lot less like hand waving if the reader has studied limits.) Using S = = 0 and = 1, so S = , and S = 2. In fact, as long as 0, 0 as well, and the infinite geometric series will have a finite sum. In this case, S = simplifies to S = which we can rewrite as S =. 2024–2025 Mathematics Instructor’s Manual 30  FORMULA FOR INFINITE GEOMETRIC SERIES An infinite geometric series equals a finite number if 0 for sufficiently large values of k. In this case, S=. EXAMPLE 2.1y: Determine the value of the infinite geometric series 2 − + − + −... SOLUTION: The initial value of this geometric sequence is 2, and r = −. Since 0 as k increases, this infinite geometric series is a finite value given by the formula S =. Therefore, S = , and S =. Arithmetic and geometric series are not the only series that have formulas for determining their values, but they are the two types of series that often occur in applied problems. We will see a few of these applications Central High School - Fort Worth, TX throughout the Mathematics Resource Guide. 2.1.3 Sigma Notation Writing out a series term by term can be very tedious, especially if there are many terms in the series. Even if we write out the series using an ellipsis, as in 3 + 7 + 11 + 15 + … + 399 + 403 + 407, we still don’t know how many terms are in the series, a piece of information we usually would like to have. Recall our notation for sequences: {xi = 4i − 1} means start with i = 1 and then generate terms by increasing the value of the index. Therefore, {xi = 4i − 1} represents the sequence 3, 7, 11, 15, …. Note that this sequence goes on forever, whereas the series 3 + 7 + 11 + 15 + … + 399 + 403 + 407 does not. We determined there were 102 terms in this sequence, so {xi = 4i − 1} generates the sequence that corresponds to this series. But how do we denote that we are interested in the series as opposed to the sequence? NOTATION Mathematicians use sigma notation, ∑ , to denote a series. means the series that corresponds to the sequence generated by the formula where the index begins at a and ends at b. For example, to express the series 3 + 7 + 11 + 15 + … + 399 + 403 + 407 in sigma notation, we write. This notation says to create the sequence given by , and then add all the terms together to create the series. Sometimes parentheses are used around the formula that generates the terms to avoid potential confusion, as in. 2024–2025 Mathematics Instructor’s Manual 31  EXAMPLE 2.1z: Write the following series in sigma notation: 1 + 4 + 9 + 16 + … + 196 + 225 + 256. SOLUTION: The terms of this series are consecutive squares, so. The first term is , and the last term is. We write. EXAMPLE 2.1aa: Determine the value of. SOLUTION: This series has 10 terms, since k is going from 1 to 10. The sequence is 1, , , , , , , , 7,381 , , so the corresponding series is the sum of these terms, which equals 2,520. Sigma notation also gives us a potentially nicer way to write the formulas for arithmetic and geometric series. Central High School - Fort Worth, TX FORMULA FOR ARITHMETIC SERIES [SIGMA FORM] FORMULA FOR FINITE GEOMETRIC SERIES [SIGMA FORM] FORMULA FOR INFINITE GEOMETRIC SERIES [SIGMA FORM] If for sufficiently large i,. Don’t be confused by the use of i and k in these equations. Here i is being used as the index, and k is being used as the last value of the index in the series. Since these formulas represent generic series, the number of terms is unknown and must be represented by a variable that is different from the variable representing the index. Any time there is some confusion about a series, especially in sigma notation, writing out a few terms is always a 2024–2025 Mathematics Instructor’s Manual 32 good idea to get a feel for what is going on. For example, the arithmetic series formula in sigma form may be particularly troublesome. What exactly is going on with ? Let’s write out a few terms. When i = 1, a + (i − 1) × d = a, so x1 = a. When i = 2, a + (i − 1) × d = a + d, so x2 = a + d. When i = 3, a + (i − 1) × d = a + 2d, so x3 = a + 2d. When i = k, a + (i − 1) × d = a + (k − 1) × d, so = a + (k − 1) × d. So this series is a + (a + d) + (a + 2d) + (a + 3d) +…+ (a + (k − 1) × d), the generic arithmetic series with k terms. Sigma notation is an extremely powerful tool that mathematicians use to write out complicated and long sums in nice shorthand. Mastering sigma notation takes a lot of time and practice. If you remember what sigma notation means, you can always write out enough terms to get a feel for what is going on. Now that we have some idea of how to use sigma notation, we will try to utilize sigma notation in our next topic: polynomials. 2.2 POLYNOMIALS Polynomials are a topic most students encounter in high school mathematics, but perhaps only briefly. Most Central High School - Fort Worth, TX polynomials encountered in high school math courses are either of degree 1 (linear) or degree 2 (quadratic). Usually degree 3 (cubics) polynomials are introduced, but higher degrees are not usually discussed much, if at all. Polynomials are also usually studied in the context of functions rather than as entities unto themselves. This means that when we are presented with a polynomial such as , we typically think of this as f(x) = and concern ourselves with the value of the function for different values of x, the graph of the function, the roots of the function, how the factored form of the function relates to the roots, and other questions about the properties of the function. Rather than considering questions such as these, here we will focus on the polynomial itself. A polynomial is an algebraic object consisting of terms. Each term is made up of a variable, usually x, raised to a different non-negative power and a coefficient, possibly 1 or 0. By convention, the polynomial is written with the powers of x in order, either from highest to lowest or lowest to highest, whichever is more convenient. The highest power of x with a non-zero coefficient is called the degree of the polynomial. The following are examples of polynomials and their degrees: ; degree 2 ; degree 3 ; degree 5 ; degree 5 ; degree 7 In general, a polynomial looks sort of like a series, but with powers of x attached to each term. For the sake of notation, we start numbering our terms with i = 0 as opposed to i = 1 as is usually done with series. This is helpful because then the index matches the exponent for the variable. A general polynomial is therefore written as. 2024–2025 Mathematics Instructor’s Manual 33 Most polynomials will not have a nice sequence of a0, a1, a2, etc., as the coefficients in a polynomial rarely follow a pattern that can be represented in a mathematical formula. So, although the sequence a0, a1, a2 … consists of “random” values, we can still think of a polynomial as a series. If it can be thought of as a series, we can write it in sigma form. SIGMA REPRESENTATION OF A POLYNOMIAL A polynomial of degree k can be written in sigma form as. Again, if this is confusing, writing out a few terms will help us see what is going on. When i = 0, ai = a0 × , so the first term is a0. When i = 1, ai = a1 × , so the second term is. When i = 2, ai = a2 × , so the third term is. When i = k, ai = ak × , so the kth term is. Central High School - Fort Worth, TX Since this is a series, we add these terms together, and our result is +…+ , our generic polynomial. 2.2.1 Adding and Subtracting Polynomials When polynomials are added and subtracted, we use an idea that in high school mathematics is generally called “combining like terms,” a particularly vague and potentially confusing phrase. What exactly is a “like term”? How do we combine them? Why do we only combine like terms, and what is wrong with combining unlike terms? These ideas are usually not given adequate or rigorous answers in most high school math courses, and so we will attempt here to address these questions mathematically using the series representation of a polynomial. Suppose we have two different arrays of numbers that we wish to add together, like two different rows in a spreadsheet. Adding the rows together should result in a new row where each entry in the row consists of the sum of the corresponding entries in the original rows. The first entry should be the sum of the two original first entries, the second entry should be the sum of the two original second entries, and so on. If we are adding two sequences together, the same idea should apply. EXAMPLE 2.2a: What sequence results if the following two sequences are added: (3, 7, 11, 15, 19, … ) + (1, 4, 9, 16, 25, … )? SOLUTION: The new sequence will