Precalculus Unit 8 PDF

Unit 8: Principle of Mathematical Induction Lesson 8.1 Proving Identities Using Mathematical Induction Contents Introduction 1 Learning Objectives 2 Warm Up...

Unit 8: Principle of Mathematical Induction Lesson 8.1 Proving Identities Using Mathematical Induction Contents Introduction 1 Learning Objectives 2 Warm Up 2 Learn about It! 3 Identity Equation 4 Principle of Mathematical Induction 5 Key Points 18 Check Your Understanding 19 Challenge Yourself 20 Bibliography 20 Key to Try It! 21 Unit 8: Principle of Mathematical Induction Lesson 8.1 Proving Identities Using Mathematical Induction Introduction How do lawyers prove the guilt of a suspect? How do judges weigh the evidences and proofs submitted by lawyers to be able to give an unbiased judgment? In the field of law, experts follow certain principles to prove an argument. In mathematics, mathematicians follow a different principle in proving a theorem, statement, or proposition. You were already familiar with proving geometric relationships, simple equations, and others that you have learned from the previous grade levels. In this lesson, you will learn a different 8.1. Proving Identities Using Mathematical Induction 1 Unit 8: Principle of Mathematical Induction method in proving statements, especially summation identities, following the principle of mathematical induction. Learning Objectives DepEd Competencies In this lesson, you should be able to do the Illustrate the Principle of following: Mathematical Induction Illustrate the principle of mathematical (STEM_PC11SMI-Ih-4). induction. Apply mathematical induction in Prove identities by applying the principle proving identities of mathematical induction. (STEM_PC11SMI-Ih-i-1). Solve problems using mathematical induction (STEM_PC11SMI-Ij-2). Warm Up Equal or Not? 10 minutes This activity will illustrate a summation identity. Materials pen paper Procedure 1. This activity will be done individually. 𝑛(𝑛+1) 2. The sum of the first 𝑛 terms of the sequence 1, 2, 3, 4, … is given by 𝑆𝑛 =. Express 2 the sum of the first 𝑛 terms of the sequence. 3. Based on the given statement, complete the table as follows. 8.1. Proving Identities Using Mathematical Induction 2 Unit 8: Principle of Mathematical Induction Data Table Table 8.1.1. Sum of the first 𝑛 terms of the sequence 1, 2, 3, 4, … Sum of the Sum of the first 𝒏 terms using the 𝒏 First 𝒏 terms 𝒏(𝒏+𝟏) first 𝒏 terms formula 𝑺𝒏 = 𝟐 1 2 3 4 5 Guide Questions 1. What can you say about the sum of the first 𝑛 terms of the sequence and their sum 𝑛(𝑛+1) using the formula 𝑆𝑛 = ? 2 2. Will your observation still be true for 𝑛 = 6? 𝑛(𝑛+1) 3. Do you think the equation 1 + 2 + 3 + ⋯ + 𝑛 = is true for any positive integer 2 greater than 6? Why or why not? Learn about It! 𝑛(𝑛+1) In the Warm Up activity, you were able to verify that 1 + 2 + 3 + ⋯ + 𝑛 = is true for the 2 first six positive integers. Let us prove similar equations by applying the principle of mathematical induction. How do we prove identity equations using the principle of mathematical induction? 8.1. Proving Identities Using Mathematical Induction 3 Unit 8: Principle of Mathematical Induction Identity Equation An identity equation is an equation that is always true for any value substituted into the 𝑛(𝑛+1) variable. Identity equations such as 1 + 2 + 3 + ⋯ + 𝑛 = can be proven using the 2 principle of mathematical induction. Some examples of identity equations are the following: 1. 2(𝑥 + 1) = 2𝑥 + 2 2. 𝑎 2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏) Example: Verify if 2(𝑥 + 1) − (𝑥 + 1)2 = −𝑥 2 + 1. 2(𝑥 + 1) − (𝑥 + 1)2 = (𝑥 + 1)(2 − 𝑥 − 1) = (𝑥 + 1)(−𝑥 + 1) = −𝑥 2 + 1 Thus, 2(𝑥 + 1) − (𝑥 + 1)2 = −𝑥 2 + 1 is an identity equation. Is 𝑥 3 − 1 = (𝑥 − 1)(𝑥 2 + 𝑥 + 1) an identity equation? 8.1. Proving Identities Using Mathematical Induction 4 Unit 8: Principle of Mathematical Induction Principle of Mathematical Induction Suppose you are in the basement of a building and you want to climb up to the 34th floor using an elevator. Will you be able to pass all the floors before your destination? Of course, before reaching the 34th floor, you will pass the first, second, third, up to the 34th floor. This is an analogy of the principle of mathematical induction. Let 𝑃(𝑛) be a statement for every integer 𝑛 from a subset of the set of integers. 𝑃(𝑛) is true for all integers 𝑛 ≥ 𝑛0 in the set if the following conditions are satisfied: 1. 𝑃(𝑛0 ) is true. 2. If 𝑃(𝑘) is true, then 𝑃(𝑘 + 1) is true for any integer 𝑘 ≥ 𝑛0. Mathematical induction can be used to prove different statements, properties, or identities. In this lesson, we will focus on proving summation identities. Let 𝑃(𝑛) be a summation identity for every natural number 𝑛. 𝑃(𝑛) is true for all 𝑛 ≥ 1 if the following conditions are satisfied: 1. 𝑃(1) is true. 2. If 𝑃(𝑘) is true, then 𝑃(𝑘 + 1) is true for any 𝑘 ≥ 1. 8.1. Proving Identities Using Mathematical Induction 5 Unit 8: Principle of Mathematical Induction Let’s Practice! Example 1 Prove that 2 + 4 + 6 + ⋯ + 2𝑛 = 𝑛(𝑛 + 1) by mathematical induction, for all positive integers 𝑛. Solution Step 1: Show that the equation is true for 𝑛 = 1. ? 2 + 4 + 6 + ⋯ + 2𝑛 = 𝑛(𝑛 + 1) ? 2 = 1(1 + 1) ? 2 = 1(2) ✓ 2=2 Hence, the equation is true for 𝑛 = 1. Step 2: Assume that the equation is true for 𝑛 = 𝑘. Then, show that it is also true for 𝑛 = 𝑘 + 1. If 𝑛 = 𝑘, we have: 2 + 4 + 6 + ⋯ + 2𝑛 = 𝑛(𝑛 + 1) 2 + 4 + 6 + ⋯ + 2𝑘 = 𝑘(𝑘 + 1) If 𝑛 = 𝑘 + 1, we have: 2 + 4 + 6 + ⋯ + 2𝑛 = 𝑛(𝑛 + 1) 2 + 4 + 6 + ⋯ + 2𝑘 + 2(𝑘 + 1) = (𝑘 + 1)[(𝑘 + 1) + 1] 2 + 4 + 6 + ⋯ + 2𝑘 + 2(𝑘 + 1) = (𝑘 + 1)(𝑘 + 2) 2 + 4 + 6 + ⋯ + 2𝑘 + 2(𝑘 + 1) = 𝑘 2 + 3𝑘 + 2 8.1. Proving Identities Using Mathematical Induction 6 Unit 8: Principle of Mathematical Induction Since 2 + 4 + 6 + ⋯ + 2𝑘 = 𝑘(𝑘 + 1), we have: 2 + 4 + 6 + ⋯ + 2𝑘 + 2(𝑘 + 1) = 𝑘(𝑘 + 1) + 2(𝑘 + 1) = 𝑘 2 + 𝑘 + 2𝑘 + 2 = 𝑘 2 + 3𝑘 + 2 Thus, by mathematical induction, the equation 2 + 4 + 6 + ⋯ + 2𝑛 = 𝑛(𝑛 + 1) is true for all positive integers 𝑛. 1 Try It! Prove that 5 + 7 + 9 + ⋯ + (2𝑛 + 3) = 𝑛(𝑛 + 4) by mathematical induction, for all positive integers 𝑛. Tips In proving summation identities, it is sometimes easier to leave one side in factored form. Looking for an expression that can be simplified easier by factoring can be helpful. For example, let us verify that 3(𝑘 + 3)3 − (𝑘 + 3)2 + 2𝑘 + 6 = (𝑘 + 3)(3𝑘 2 + 17𝑘 + 26) is an identity. It is easier to factor the left-hand side of the equation instead of expanding both sides. (𝑘 + 3)[3(𝑘 + 3)2 − (𝑘 + 3) + 2] = (𝑘 + 3)[3(𝑘 2 + 6𝑘 + 9) − 𝑘 − 3 + 2] = (𝑘 + 3)(3𝑘 2 + 18𝑘 + 27 − 𝑘 − 1) = (𝑘 + 3)(3𝑘 2 + 17𝑘 + 26) 8.1. Proving Identities Using Mathematical Induction 7 Unit 8: Principle of Mathematical Induction Example 2 𝑛(9𝑛+5) Use mathematical induction to prove 7 + 16 + 25 + ⋯ + (9𝑛 − 2) = , for all positive 2 integers 𝑛. Solution Step 1: Show that the equation is true for 𝑛 = 1. ?𝑛(9𝑛 + 5) 7 + 16 + 25 + ⋯ + (9𝑛 − 2) = 2 ? 1[9(1) + 5] 7= 2 ? 9+5 7= 2 ? 14 7= 2 ✓ 7=7 Hence, the equation is true for 𝑛 = 1. Step 2: Assume that the equation is true for 𝑛 = 𝑘. Then, show that it is also true for 𝑛 = 𝑘 + 1. If 𝑛 = 𝑘, we have: 𝑛(9𝑛 + 5) 7 + 16 + 25 + ⋯ + (9𝑛 − 2) = 2 𝑘(9𝑘 + 5) 7 + 16 + 25 + ⋯ + (9𝑘 − 2) = 2 8.1. Proving Identities Using Mathematical Induction 8 Unit 8: Principle of Mathematical Induction If 𝑛 = 𝑘 + 1, we have: 𝑛(9𝑛 + 5) 7 + 16 + 25 + ⋯ + (9𝑛 − 2) = 2 (𝑘 + 1)[9(𝑘 + 1) + 5] 7 + 16 + 25 + ⋯ + (9𝑘 − 2) + [9(𝑘 + 1) − 2] = 2 (𝑘 + 1)(9𝑘 + 9 + 5) 7 + 16 + 25 + ⋯ + (9𝑘 − 2) + (9𝑘 + 9 − 2) = 2 (𝑘 + 1)(9𝑘 + 14) 7 + 16 + 25 + ⋯ + (9𝑘 − 2) + (9𝑘 + 7) = 2 2 9𝑘 + 23𝑘 + 14 7 + 16 + 25 + ⋯ + (9𝑘 − 2) + (9𝑘 + 7) = 2 𝑘(9𝑘+5) Since 7 + 16 + 25 + ⋯ + (9𝑘 − 2) = , we have: 2 𝑘(9𝑘 + 5) 7 + 16 + 25 + ⋯ + (9𝑘 − 2) + (9𝑘 + 7) = + 9𝑘 + 7 2 9𝑘 2 + 5𝑘 + 2(9𝑘 + 7) = 2 2 9𝑘 + 5𝑘 + 18𝑘 + 14 = 2 2 9𝑘 + 23𝑘 + 14 = 2 𝑛(9𝑛+5) Thus, by mathematical induction, the equation 7 + 16 + 25 + ⋯ + (9𝑛 − 2) = is true 2 for all positive integers 𝑛. 2 Try It! 𝑛(7𝑛+5) Use mathematical induction to prove 6 + 13 + 20 + ⋯ + (7𝑛 − 1) = , for all 2 positive integers 𝑛. 8.1. Proving Identities Using Mathematical Induction 9 Unit 8: Principle of Mathematical Induction Example 3 −3+3𝑛+1 Show that 3 + 9 + 27 + ⋯ + 3𝑛 = is an identity using mathematical induction, for all 2 positive integers 𝑛. Solution Step 1: Show that the equation is true for 𝑛 = 1. ?−3 + 3𝑛+1 3 + 9 + 27 + ⋯ + 3𝑛 = 2 1+1 ? −3 + 3 3= 2 2 ? −3 + 3 3= 2 ? −3 + 9 3= 2 ? 6 3= 2 ✓ 3=3 Hence, the equation is true for 𝑛 = 1. Step 2: Assume that the equation is true for 𝑛 = 𝑘. Then, show that it is also true for 𝑛 = 𝑘 + 1. If 𝑛 = 𝑘, we have: −3 + 3𝑛+1 3 + 9 + 27 + ⋯ + 3𝑛 = 2 −3 + 3𝑘+1 3 + 9 + 27 + ⋯ + 3𝑘 = 2 8.1. Proving Identities Using Mathematical Induction 10 Unit 8: Principle of Mathematical Induction If 𝑛 = 𝑘 + 1, we have: −3 + 3𝑛+1 3 + 9 + 27 + ⋯ + 3𝑛 = 2 −3 + 3𝑘+1+1 3 + 9 + 27 + ⋯ + 3𝑘 + 3𝑘+1 = 2 −3 + 3𝑘+2 3 + 9 + 27 + ⋯ + 3𝑘 + 3𝑘+1 = 2 −3+3𝑘+1 Since 3 + 9 + 27 + ⋯ + 3𝑘 = , we have: 2 −3 + 3𝑘+1 3 + 9 + 27 + ⋯ + 3𝑘 + 3𝑘+1 = + 3𝑘+1 2 −3 + 3𝑘+1 + 2(3𝑘+1 ) = 2 𝑘+1 −3 + 3(3 ) = 2 −3 + 3𝑘+2 = 2 −3+3𝑛+1 Thus, by mathematical induction, 3 + 9 + 27 + ⋯ + 3𝑛 = is an identity for all positive 2 integers 𝑛. 3 Try It! −5+5(4𝑛 ) Show that 5 + 20 + 80 + ⋯ + 5(4𝑛−1 ) = is an identity using mathematical 3 induction, for all positive integers 𝑛. 8.1. Proving Identities Using Mathematical Induction 11 Unit 8: Principle of Mathematical Induction Example 4 𝑛(𝑛+1)(2𝑛+1) Prove that 12 + 22 + 32 + ⋯ + 𝑛2 = using mathematical induction, for all positive 6 integers 𝑛. Solution Step 1: Show that the equation is true for 𝑛 = 1. ?𝑛(𝑛 + 1)(2𝑛 + 1) 12 + 22 + 32 + ⋯ + 𝑛2 = 6 ? 1(1 + 1)[2(1) + 1] 12 = 6 ? 1(2)(3) 1= 6 ? 6 1= 6 ✓ 1=1 Hence, the equation is true for 𝑛 = 1. Step 2: Assume that the equation is true for 𝑛 = 𝑘. Then, show that it is also true for 𝑛 = 𝑘 + 1. If 𝑛 = 𝑘, we have: 𝑛(𝑛 + 1)(2𝑛 + 1) 12 + 22 + 32 + ⋯ + 𝑛2 = 6 𝑘(𝑘 + 1)(2𝑘 + 1) 12 + 22 + 32 + ⋯ + 𝑘 2 = 6 8.1. Proving Identities Using Mathematical Induction 12 Unit 8: Principle of Mathematical Induction If 𝑛 = 𝑘 + 1, we have: 𝑘(𝑘 + 1)(2𝑘 + 1) 12 + 22 + 32 + ⋯ + 𝑘 2 = 6 (𝑘 + 1)[(𝑘 + 1) + 1][2(𝑘 + 1) + 1] 12 + 22 + 32 + ⋯ + 𝑘 2 + (𝑘 + 1)2 = 6 (𝑘 + 1)(𝑘 + 2)(2𝑘 + 3) 12 + 22 + 32 + ⋯ + 𝑘 2 + (𝑘 + 1)2 = 6 3 2 2𝑘 + 9𝑘 + 13𝑘 + 6 12 + 22 + 32 + ⋯ + 𝑘 2 + (𝑘 + 1)2 = 6 𝑘(𝑘+1)(2𝑘+1) Since 12 + 22 + 32 + ⋯ + 𝑘 2 = , we have: 6 𝑘(𝑘 + 1)(2𝑘 + 1) 12 + 22 + 32 + ⋯ + 𝑘 2 + (𝑘 + 1)2 = + (𝑘 + 1)2 6 2𝑘 3 + 3𝑘 2 + 𝑘 + 6(𝑘 2 + 2𝑘 + 1) = 6 2𝑘 + 3𝑘 + 𝑘 + 6𝑘 2 + 12𝑘 + 6 3 2 = 6 3 2 2𝑘 + 9𝑘 + 13𝑘 + 6 = 6 𝑛(𝑛+1)(2𝑛+1) Thus, by mathematical induction, the equation 12 + 22 + 32 + ⋯ + 𝑛2 = is true for 6 all positive integers 𝑛. 4 Try It! 𝑛(𝑛+1)(2𝑛+7)+6𝑛 Prove that 22 + 32 + 42 + ⋯ + (𝑛 + 1)2 = using mathematical induction, 6 for all positive integers 𝑛. 8.1. Proving Identities Using Mathematical Induction 13 Unit 8: Principle of Mathematical Induction Example 5 Prove the statement below using mathematical induction, for all positive integers 𝑛. 𝑛 𝑛(𝑛 + 1)(2𝑛 + 4) ∑(𝑖 2 + 𝑖) = 6 𝑖=1 Solution Step 1: Show that the equation is true for 𝑛 = 1. 1 ? 𝑛(𝑛 + 1)(2𝑛 + 4) ∑(𝑖 2 + 𝑖) = 6 𝑖=1 1(1 + 1)[2(1) + 4] ? 12 + 1 = 6 ? 1(2)(6) 1+1 = 6 ✓ 2=2 Hence, the equation is true for 𝑛 = 1. Step 2: Assume that the equation is true for 𝑛 = 𝑘. Then, show that it is also true for 𝑛 = 𝑘 + 1. If 𝑛 = 𝑘, we have: 𝑛 𝑛(𝑛 + 1)(2𝑛 + 4) ∑(𝑖 2 + 𝑖) = 6 𝑖=1 𝑘 𝑘(𝑘 + 1)(2𝑘 + 4) ∑(𝑖 2 + 𝑖) = 6 𝑖=1 8.1. Proving Identities Using Mathematical Induction 14 Unit 8: Principle of Mathematical Induction If 𝑛 = 𝑘 + 1, we have: 𝑘 (𝑘 + 1)(𝑘 + 1 + 1)[2(𝑘 + 1) + 4] ∑(𝑖 2 + 𝑖) + (𝑘 + 1)2 + (𝑘 + 1) = 6 𝑖=1 𝑘 (𝑘 + 1)(𝑘 + 2)(2𝑘 + 2 + 4) ∑(𝑖 2 + 𝑖) + (𝑘 + 1)2 + 𝑘 + 1 = 6 𝑖=1 𝑘 (𝑘 2 + 3𝑘 + 2)(2𝑘 + 6) ∑(𝑖 2 + 𝑖) + 𝑘 2 + 2𝑘 + 1 + 𝑘 + 1 = 6 𝑖=1 𝑘 2𝑘 3 + 12𝑘 2 + 22𝑘 + 12 ∑(𝑖 2 + 𝑖) + 𝑘2 + 3𝑘 + 2 = 6 𝑖=1 𝑘 𝑘 3 + 6𝑘 2 + 11𝑘 + 6 ∑(𝑖 2 + 𝑖) + 𝑘 2 + 3𝑘 + 2 = 3 𝑖=1 𝑘(𝑘+1)(2𝑘+4) Since ∑𝑘𝑖=1(𝑖 2 + 𝑖) = , we have: 6 𝑘 𝑘(𝑘 + 1)(2𝑘 + 4) ∑(𝑖 2 + 𝑖) + 𝑘 2 + 3𝑘 + 2 = + 𝑘 2 + 3𝑘 + 2 6 𝑖=1 2𝑘 3 + 6𝑘 2 + 4𝑘 = + 𝑘 2 + 3𝑘 + 2 6 2𝑘 3 + 6𝑘 2 + 4𝑘 + 6(𝑘 2 + 3𝑘 + 2) = 6 2𝑘 + 6𝑘 + 4𝑘 + 6𝑘 2 + 18𝑘 + 12 3 2 = 6 3 2 2𝑘 + 12𝑘 + 22𝑘 + 12 = 6 3 2 𝑘 + 6𝑘 + 11𝑘 + 6 = 3 𝑛(𝑛+1)(2𝑛+4) Thus, by mathematical induction, the statement ∑𝑛𝑖=1(𝑖 2 + 𝑖) = is true for all 6 positive integers 𝑛. 8.1. Proving Identities Using Mathematical Induction 15 Unit 8: Principle of Mathematical Induction 5 Try It! Prove the equation below using mathematical induction, for all positive integers 𝑛. 𝑛 𝑛(𝑛 + 1)(2𝑛 − 11) ∑(𝑖 − 2)2 = + 4𝑛 6 𝑖=1 Example 6 Prove by mathematical induction that the sum of the first 𝑛 terms of the sequence 1 1 1 1 , , ,… is 𝑆𝑛 = 1 − 𝑛+1. 1(2) 2(3) 3(4) Solution Step 1: Determine the 𝑛th term of the sequence. 1 1 1 Let 𝑎𝑛 be the 𝑛th term of the sequence 1(2) , 2(3) , 3(4) , …. The sequence can be 1 1 1 1 rewritten as 1(1+1) , 2(2+1) , 3(3+1) , …. Hence, 𝑎𝑛 = 𝑛(𝑛+1). 1 1 1 1 1 We have to prove that 1(2) + + + ⋯ + 𝑛(𝑛+1) = 1 − 𝑛+1. 2(3) 3(4) Step 2: Show that the equation is true for 𝑛 = 1. 1 1 1 1 ? 1 + + +⋯+ = 1− 1(2) 2(3) 3(4) 𝑛(𝑛 + 1) 𝑛+1 1 ? 1 = 1− 2 1+1 1 ? 1 = 1− 2 2 1✓ 1 = 2 2 Hence, the equation is true for 𝑛 = 1. 8.1. Proving Identities Using Mathematical Induction 16 Unit 8: Principle of Mathematical Induction Step 3: Assume that the equation is true for 𝑛 = 𝑘, then show that it is also true for 𝑛 = 𝑘 + 1. If 𝑛 = 𝑘, we have: 1 1 1 1 1 + + + ⋯+ =1− 1(2) 2(3) 3(4) 𝑛(𝑛 + 1) 𝑛+1 1 1 1 1 1 + + +⋯+ =1− 1(2) 2(3) 3(4) 𝑘(𝑘 + 1) 𝑘+1 If 𝑛 = 𝑘 + 1, we have: 1 1 1 1 1 1 + + + ⋯+ + =1− 1(2) 2(3) 3(4) 𝑘(𝑘 + 1) (𝑘 + 1)(𝑘 + 1 + 1) 𝑘+1+1 1 1 1 1 1 1 + + + ⋯+ + =1− 1(2) 2(3) 3(4) 𝑘(𝑘 + 1) (𝑘 + 1)(𝑘 + 2) 𝑘+2 1 1 1 1 1 𝑘+2−1 + + + ⋯+ + = 1(2) 2(3) 3(4) 𝑘(𝑘 + 1) (𝑘 + 1)(𝑘 + 2) 𝑘+2 1 1 1 1 1 𝑘+1 + + + ⋯+ + = 1(2) 2(3) 3(4) 𝑘(𝑘 + 1) (𝑘 + 1)(𝑘 + 2) 𝑘 + 2 1 1 1 1 1 Since 1(2) + + + ⋯ + 𝑘(𝑘+1) = 1 − 𝑘+1, we have: 2(3) 3(4) 1 1 1 1 1 1 1 + + + ⋯+ + =1− + 1(2) 2(3) 3(4) 𝑘(𝑘 + 1) (𝑘 + 1)(𝑘 + 2) 𝑘 + 1 (𝑘 + 1)(𝑘 + 2) 𝑘+1−1 1 = + 𝑘+1 (𝑘 + 1)(𝑘 + 2) 𝑘 1 = + 𝑘 + 1 (𝑘 + 1)(𝑘 + 2) 𝑘(𝑘 + 2) + 1 = (𝑘 + 1)(𝑘 + 2) 𝑘 2 + 2𝑘 + 1 = (𝑘 + 1)(𝑘 + 2) (𝑘 + 1)(𝑘 + 1) = (𝑘 + 1)(𝑘 + 2) 𝑘+1 = 𝑘+2 8.1. Proving Identities Using Mathematical Induction 17 Unit 8: Principle of Mathematical Induction 1 1 1 1 1 Thus, by mathematical induction, the equation 1(2) + + + ⋯ + 𝑛(𝑛+1) = 1 − 𝑛+1 is true 2(3) 3(4) for all positive integers 𝑛. 6 Try It! Prove by mathematical induction that the sum of the first 𝑛 terms of the sequence 1 1 1 1 1 , , ,… is 𝑆𝑛 = 2 − 𝑛+2. 2(3) 3(4) 4(5) Key Points ___________________________________________________________________________________________ An identity equation is an equation that is always true for any value substituted into the variable. Let 𝑃(𝑛) be a summation identity for every natural number 𝑛. 𝑃(𝑛) is true for all 𝑛 ≥ 1 if the following conditions are satisfied: ○ 𝑃(1) is true. ○ If 𝑃(𝑘) is true, then 𝑃(𝑘 + 1) is also true for all 𝑘 ≥ 1. ___________________________________________________________________________________________ 8.1. Proving Identities Using Mathematical Induction 18 Unit 8: Principle of Mathematical Induction Check Your Understanding A. Write TRUE if the statement is correct. Otherwise, write FALSE. 1. The equation (𝑥 + 1)2 (𝑥 − 2) = 𝑥 3 + 2𝑥 2 − 5𝑥 + 2 is an identity. 2. The equation (𝑥 − 1)3 (2𝑥 + 1)2 = (2𝑥 2 − 𝑥 − 1)(𝑥 − 1)(𝑥 2 + 𝑥 + 1) is an identity. 3. If an equation holds true for 𝑛 = 1, then it is also true for 𝑛 = 2. 4. Mathematical induction can be used to prove identities for any real number greater than or equal to one. 5. The equation 4 + 8 + 12 + ⋯ + 4𝑛 = 3𝑛2 + 1 holds true for 𝑛 = 1. B. Prove by mathematical induction that the following equations are true for all positive integers 𝑛. 1. 2 + 8 + 14 + ⋯ + (6𝑛 − 4) = 𝑛(3𝑛 − 1) 𝑛(7𝑛+3) 2. 5 + 12 + 19 + ⋯ + (7𝑛 − 2) = 2 𝑛(11𝑛−47) 3. −18 − 7 + 4 + ⋯ + (11𝑛 − 29) = 2 −4+4𝑛+1 4. 4 + 16 + 64 + ⋯ + 4𝑛 = 3 7 5. 7 + 21 + 63 + ⋯ + 7(3𝑛−1 ) = − 2 (1 − 3𝑛 ) 1 1 1 2𝑛−1 6. + 8 + 4 + ⋯ + 2𝑛−5 = 16 16 (𝑛 2+𝑛)(2𝑛+1)−18𝑛 7. ∑𝑛𝑖=1(𝑖 2 − 3) = 6 𝑛(2𝑛 2+3𝑛+1) 8. ∑𝑛𝑖=1[(𝑖 − 2)2 + 4𝑖] = + 4𝑛 6 𝑛 2 (𝑛+1)2 9. ∑𝑛𝑖=1 𝑖 3 = 4 𝑛 2 (𝑛+1)2 +4𝑛 10. ∑𝑛𝑖=1(𝑖 3 + 1) = 4 8.1. Proving Identities Using Mathematical Induction 19 Unit 8: Principle of Mathematical Induction Challenge Yourself Prove by mathematical induction that the sum of the first 𝑛 terms of the given sequence is equal to 𝑆𝑛. −72(2𝑛−3𝑛) 1. 36, 54, 81, … ; 𝑆𝑛 = 2𝑛 𝑛(2𝑎𝑛−5𝑛+9) 2. (𝑎 + 2), (3𝑎 − 3), (5𝑎 − 8), … ; 𝑆𝑛 = 2 1 1 1 1 1 3. , , , … ; 𝑆𝑛 = − 12 20 30 3 𝑛+3 1 1 1 3𝑛 2+5𝑛 4. , , , … ; 𝑆𝑛 = 4(𝑛+1)(𝑛+2) 3 8 15 1 2 1 5𝑛 2+13𝑛 5. , , ,…; 𝑆𝑛 = 6(𝑛+2)(𝑛+3) 4 15 12 Bibliography Barnett, Raymond, Michael Ziegler, Karl Byleen, and David Sobecki. College Algebra with Trigonometry. Boston: McGraw Hill Higher Education, 2008. Bittinger, Marvin L., Judith A. Beecher, David J. Ellenbogen, and Judith A. Penna. Algebra and Trigonometry: Graphs and Models. 4th ed. Boston: Pearson/Addison Wesley, 2009. Blitzer, Robert. Algebra and Trigonometry. 3rd ed. Upper Saddle River, New Jersey: Pearson/Prentice Hall, 2007. Larson, Ron. College Algebra with Applications for Business and the Life Sciences. Boston: MA:Houghton Mifflin, 2009. Simmons, George F. Calculus with Analytic Geometry. 2nd ed. New York: McGraw-Hill, 1996. 8.1. Proving Identities Using Mathematical Induction 20 Unit 8: Principle of Mathematical Induction Key to Try It! 1. Prove that 5 + 7 + 9 + ⋯ + (2𝑛 + 3) = 𝑛(𝑛 + 4) by mathematical induction, for all positive integers 𝑛. Step 1: Show that the equation is true for 𝑛 = 1. ? 5 = 1(1 + 4) ✓ 5=5 Step 2: Assume that the equation is true for 𝑛 = 𝑘. Then, show that it is also true for 𝑛 = 𝑘 + 1. For 𝑛 = 𝑘, we have: 5 + 7 + 9 + ⋯ + (2𝑛 + 3) = 𝑛(𝑛 + 4) 5 + 7 + 9 + ⋯ + (2𝑘 + 3) = 𝑘(𝑘 + 4) For 𝑛 = 𝑘 + 1, we have: 5 + 7 + 9 + ⋯ + (2𝑘 + 3) + [2(𝑘 + 1) + 3] = (𝑘 + 1)(𝑘 + 1 + 4) 5 + 7 + 9 + ⋯ + (2𝑘 + 3) + (2𝑘 + 5) = (𝑘 + 1)(𝑘 + 5) 5 + 7 + 9 + ⋯ + (2𝑘 + 3) + (2𝑘 + 5) = 𝑘 2 + 6𝑘 + 5 Since 5 + 7 + 9 + ⋯ + (2𝑘 + 3) = 𝑘(𝑘 + 4), we have: 5 + 7 + 9 + ⋯ + (2𝑘 + 3) + (2𝑘 + 5) = 𝑘(𝑘 + 4) + 2𝑘 + 5 = 𝑘 2 + 4𝑘 + 2𝑘 + 5 = 𝑘 2 + 6𝑘 + 5 Thus, by mathematical induction, the equation 5 + 7 + 9 + ⋯ + (2𝑛 + 3) = 𝑛(𝑛 + 4) is true for all positive integers 𝑛. 8.1. Proving Identities Using Mathematical Induction 21 Unit 8: Principle of Mathematical Induction 𝑛(7𝑛+5) 2. Use mathematical induction to prove 6 + 13 + 20 + ⋯ + (7𝑛 − 1) = , for all 2 positive integers 𝑛. Step 1: Show that the equation is true for 𝑛 = 1. 1[7(1) + 5] ? 6= 2 ? 12 6= 2 ✓ 6=6 Step 2: Assume that the equation is true for 𝑛 = 𝑘. Then, show that it is also true for 𝑛 = 𝑘 + 1. For 𝑛 = 𝑘, we have: 𝑛(7𝑛 + 5) 6 + 13 + 20 + ⋯ + (7𝑛 − 1) = 2 𝑘(7𝑘 + 5) 6 + 13 + 20 + ⋯ + (7𝑘 − 1) = 2 For 𝑛 = 𝑘 + 1, we have: (𝑘 + 1)[7(𝑘 + 1) + 5] 6 + 13 + 20 + ⋯ + (7𝑘 − 1) + [7(𝑘 + 1) − 1] = 2 (𝑘 + 1)(7𝑘 + 12) 6 + 13 + 20 + ⋯ + (7𝑘 − 1) + (7𝑘 + 6) = 2 2 7𝑘 + 19𝑘 + 12 6 + 13 + 20 + ⋯ + (7𝑘 − 1) + (7𝑘 + 6) = 2 8.1. Proving Identities Using Mathematical Induction 22 Unit 8: Principle of Mathematical Induction 𝑘(7𝑘+5) Since 6 + 13 + 20 + ⋯ + (7𝑘 − 1) = , we have: 2 𝑘(7𝑘 + 5) 6 + 13 + 20 + ⋯ + (7𝑘 − 1) + (7𝑘 + 6) = + (7𝑘 + 6) 2 7𝑘 2 + 5𝑘 + 14𝑘 + 12 = 2 2 7𝑘 + 19𝑘 + 12 = 2 𝑛(7𝑛+5) Thus, by mathematical induction, the equation 6 + 13 + 20 + ⋯ + (7𝑛 − 1) = 2 is true for all positive integers 𝑛. −5+5(4𝑛 ) 3. Show that 5 + 20 + 80 + ⋯ + 5(4𝑛−1 ) = is an identity using mathematical 3 induction, for all positive integers 𝑛. Step 1: Show that the equation is true for 𝑛 = 1. −5 + 5(41 ) ? 5= 3 ? 15 5= 3 ✓ 5=5 Step 2: Assume that the equation is true for 𝑛 = 𝑘. Then, show that it is also true for 𝑛 = 𝑘 + 1. For 𝑛 = 𝑘, we have: −5 + 5(4𝑛 ) 5 + 20 + 80 + ⋯ + 5(4𝑛−1 ) = 3 −5 + 5(4𝑘 ) 5 + 20 + 80 + ⋯ + 5(4𝑘−1 ) = 3 8.1. Proving Identities Using Mathematical Induction 23 Unit 8: Principle of Mathematical Induction For 𝑛 = 𝑘 + 1, we have: −5 + 5(4𝑘+1 ) 5 + 20 + 80 + ⋯ + 5(4𝑘−1 ) + 5(4𝑘+1−1 ) = 3 −5 + 5(4𝑘+1 ) 5 + 20 + 80 + ⋯ + 5(4𝑘−1 ) + 5(4𝑘 ) = 3 −5+5(4𝑘 ) Since 5 + 20 + 80 + ⋯ + 5(4𝑘−1 ) = , we have: 3 −5 + 5(4𝑘 ) 5 + 20 + 80 + ⋯ + 5(4𝑘−1 ) + 5(4𝑘 ) = + 5(4𝑘 ) 3 −5 + 5(4𝑘 ) + 15(4𝑘 ) = 3 −5 + 20(4𝑘 ) = 3 −5 + 5(4)(4𝑘 ) = 3 −5 + 5(4𝑘+1 ) = 3 −5+5(4𝑛 ) Thus, by mathematical induction, the equation 5 + 20 + 80 + ⋯ + 5(4𝑛−1 ) = is 3 true for all positive integers 𝑛. 𝑛(𝑛+1)(2𝑛+7)+6𝑛 4. Prove that 22 + 32 + 42 + ⋯ + (𝑛 + 1)2 = using mathematical induction, 6 for all positive integers 𝑛. Step 1: Show that the equation is true for 𝑛 = 1. 1(1 + 1)[2(1) + 7] + 6(1) ? 22 = 6 ? 2(9) + 6 4= 6 ? 24 4= 6 ✓ 4=4 8.1. Proving Identities Using Mathematical Induction 24 Unit 8: Principle of Mathematical Induction Step 2: Assume that the equation is true for 𝑛 = 𝑘. Then, show that it is also true for 𝑛 = 𝑘 + 1. For 𝑛 = 𝑘, we have: 𝑛(𝑛 + 1)(2𝑛 + 7) + 6𝑛 22 + 32 + 42 + ⋯ + (𝑛 + 1)2 = 6 𝑘(𝑘 + 1)(2𝑘 + 7) + 6𝑘 22 + 32 + 42 + ⋯ + (𝑘 + 1)2 = 6 For 𝑛 = 𝑘 + 1, we have: (𝑘 + 1)(𝑘 + 1 + 1)[2(𝑘 + 1) + 7] + 6(𝑘 + 1) 22 + 32 + 42 + ⋯ + (𝑘 + 1)2 + (𝑘 + 1 + 1)2 = 6 (𝑘 + 1)(𝑘 + 2)(2𝑘 + 9) + 6(𝑘 + 1) 22 + 32 + 42 + ⋯ + (𝑘 + 1)2 + (𝑘 + 2)2 = 6 3 2 2𝑘 + 15𝑘 + 31𝑘 + 18 + 6𝑘 + 6 22 + 32 + 42 + ⋯ + (𝑘 + 1)2 + (𝑘 2 + 4𝑘 + 4) = 6 3 2 2𝑘 + 15𝑘 + 37𝑘 + 24 22 + 32 + 42 + ⋯ + (𝑘 + 1)2 + (𝑘 2 + 4𝑘 + 4) = 6 𝑘(𝑘+1)(2𝑘+7)+6𝑘 Since 22 + 32 + 42 + ⋯ + (𝑘 + 1)2 = , we have: 6 𝑘(𝑘 + 1)(2𝑘 + 7) + 6𝑘 22 + 32 + 42 + ⋯ + (𝑘 + 1)2 + (𝑘 2 + 4𝑘 + 4) = + 𝑘 2 + 4𝑘 + 4 6 2𝑘 3 + 9𝑘 2 + 7𝑘 + 6𝑘 + 6(𝑘 2 + 4𝑘 + 4) = 6 2𝑘 + 9𝑘 + 13𝑘 + 6𝑘 2 + 24𝑘 + 24 3 2 = 6 3 2 2𝑘 + 15𝑘 + 37𝑘 + 24 = 6 Thus, by mathematical induction, the equation 𝑛(𝑛+1)(2𝑛+7)+6𝑛 22 + 32 + 42 + ⋯ + (𝑛 + 1)2 = is true for all positive integers 𝑛. 6 8.1. Proving Identities Using Mathematical Induction 25 Unit 8: Principle of Mathematical Induction 5. Prove the statement below using mathematical induction, for all positive integers 𝑛. 𝑛 𝑛(𝑛 + 1)(2𝑛 − 11) ∑(𝑖 − 2)2 = + 4𝑛 6 𝑖=1 Step 1: Show that the equation is true for 𝑛 = 1. 1 ? 1(1 + 1)[2(1) − 11] ∑(𝑖 − 2)2 = + 4(1) 6 𝑖=1 ? 2(−9) (1 − 2)2 = +4 6 ? 1 =− 3 + 4 ✓ 1=1 Step 2: Assume that the equation is true for 𝑛 = 𝑘. Then, show that it is also true for 𝑛 = 𝑘 + 1. For 𝑛 = 𝑘, we have: 𝑛 𝑛(𝑛 + 1)(2𝑛 − 11) ∑(𝑖 − 2)2 = + 4𝑛 6 𝑖=1 𝑘 𝑘(𝑘 + 1)(2𝑘 − 11) ∑(𝑖 − 2)2 = + 4𝑘 6 𝑖=1 For 𝑛 = 𝑘 + 1, we have: 𝑘 (𝑘 + 1)(𝑘 + 1 + 1)[2(𝑘 + 1) − 11] ∑(𝑖 − 2)2 + (𝑘 + 1 − 2)2 = + 4(𝑘 + 1) 6 𝑖=1 𝑘 (𝑘 + 1)(𝑘 + 2)(2𝑘 − 9) + 24(𝑘 + 1) ∑(𝑖 − 2)2 + (𝑘 − 1)2 = 6 𝑖=1 𝑘 2𝑘 3 − 3𝑘 2 + 𝑘 + 6 ∑(𝑖 − 2)2 + (𝑘 − 1)2 = 6 𝑖=1 8.1. Proving Identities Using Mathematical Induction 26 Unit 8: Principle of Mathematical Induction 𝑘(𝑘+1)(2𝑘−11) Since ∑𝑘𝑖=1(𝑖 − 2)2 = + 4𝑘, we have: 6 𝑘 𝑘(𝑘 + 1)(2𝑘 − 11) ∑(𝑖 − 2)2 + (𝑘 − 1)2 = + 4𝑘 + (𝑘 − 1)2 6 𝑖=1 𝑘(𝑘 + 1)(2𝑘 − 11) + 24𝑘 + 6(𝑘 − 1)2 = 6 2𝑘 − 9𝑘 − 11𝑘 + 24𝑘 + 6𝑘 2 − 12𝑘 + 6 3 2 = 6 3 2 2𝑘 − 3𝑘 + 𝑘 + 6 = 6 𝑛(𝑛+1)(2𝑛−11) Thus, by mathematical induction, the equation ∑𝑛𝑖=1(𝑖 − 2)2 = + 4𝑛 is true 6 for all positive integers 𝑛. 6. Prove by mathematical induction that the sum of the first 𝑛 terms of the sequence 1 1 1 1 1 , , ,… is 𝑆𝑛 = 2 − 𝑛+2. 2(3) 3(4) 4(5) Step 1: Show that the equation is true for 𝑛 = 1. 𝑛 1 ? 1 1 ∑ = − (𝑖 + 1)(𝑖 + 2) 2 𝑛 + 2 𝑖=1 1 1 ? 1 1 ∑[ ]= − (𝑖 + 1)(𝑖 + 2) 2 1+2 𝑖=1 1 ? 1 1 = − (1 + 1)(1 + 2) 2 3 1 ? 1 = 2(3) 6 1✓1 = 6 6 Step 2: Assume that the equation is true for 𝑛 = 𝑘. Then, show that it is also true for 𝑛 = 𝑘 + 1. 8.1. Proving Identities Using Mathematical Induction 27 Unit 8: Principle of Mathematical Induction For 𝑛 = 𝑘, we have: 𝑛 1 1 1 ∑[ ]= − (𝑖 + 1)(𝑖 + 2) 2 𝑛+2 𝑖=1 𝑘 1 1 1 ∑[ ]= − (𝑖 + 1)(𝑖 + 2) 2 𝑘+2 𝑖=1 For 𝑛 = 𝑘 + 1, we have: 𝑘 1 1 1 1 ∑[ ]+ = − (𝑖 + 1)(𝑖 + 2) (𝑘 + 1 + 1)(𝑘 + 1 + 2) 2 𝑘 + 1 + 2 𝑖=1 𝑘 1 1 1 1 ∑[ ]+ = − (𝑖 + 1)(𝑖 + 2) (𝑘 + 2)(𝑘 + 3) 2 𝑘 + 3 𝑖=1 𝑘 1 1 𝑘+3−2 ∑[ ]+ = (𝑖 + 1)(𝑖 + 2) (𝑘 + 2)(𝑘 + 3) 2(𝑘 + 3) 𝑖=1 𝑘 1 1 𝑘+1 ∑[ ]+ = (𝑖 + 1)(𝑖 + 2) (𝑘 + 2)(𝑘 + 3) 2(𝑘 + 3) 𝑖=1 1 1 1 Since ∑𝑘𝑖=1 [(𝑖+1)(𝑖+2)] = 2 − 𝑘+2, we have: 𝑘 1 1 1 1 1 ∑[ ]+ = − + (𝑖 + 1)(𝑖 + 2) (𝑘 + 2)(𝑘 + 3) 2 𝑘 + 2 (𝑘 + 2)(𝑘 + 3) 𝑖=1 (𝑘 + 2)(𝑘 + 3) − 2(𝑘 + 3) + 2 = 2(𝑘 + 2)(𝑘 + 3) 𝑘 2 + 5𝑘 + 6 − 2𝑘 − 6 + 2 = 2(𝑘 + 2)(𝑘 + 3) 𝑘 2 + 3𝑘 + 2 = 2(𝑘 + 2)(𝑘 + 3) (𝑘 + 2)(𝑘 + 1) = 2(𝑘 + 2)(𝑘 + 3) 𝑘+1 = 2(𝑘 + 3) 1 1 1 Thus, by mathematical induction, the equation ∑𝑛𝑖=1 (𝑖+1)(𝑖+2) = 2 − 𝑛+2 is true for all positive integers 𝑛. 8.1. Proving Identities Using Mathematical Induction 28