Topic 1 Binomial Expansion and Hayah PDF

Topic 1: Binomial Expansion and Hayah TOPIC 1: BINOMIAL EXPANSION AND HAYAH 1.1 Binomial Theorem When n Is A Positive Integer LEARNING OUTCOMES At the end of the lecture, students should be able to: Expand ( a + b ) where n is a positive integer...

Topic 1: Binomial Expansion and Hayah TOPIC 1: BINOMIAL EXPANSION AND HAYAH 1.1 Binomial Theorem When n Is A Positive Integer LEARNING OUTCOMES At the end of the lecture, students should be able to: Expand ( a + b ) where n is a positive integer n n Write n! notations and nCr =   as a binomial coefficient r Determine the general term in a binomial expansion ( a + b ) n where n is a positive integer Find Tr +1 term of the binomial expansion. An expression such as ( a + x ) which consists of two terms is called a binomial expression and (a + x)n is a binomial function. Let consider the expansion of (a + x)n. ( a + x) 0 = 1 (a + x)1 = a+x (a + x) 2 = a 2 + 2ax + x 2 ( a + x)3 = a 3 + 3a 2 x + 3ax 2 + x3 (a + x) 4 = a 4 + 4a 3 x + 6a 2 x 2 + 4ax3 + x 4 (a + x)5 = a 5 + 5a 4 x + 10a 3 x 2 + 10a 2 x3 + 5ax 4 + x5 Now, ignore the numerical ∗coefficient ( the numbers 2, 3, 4, 5, 6 and 10) and let consider only on the remaining constant 𝛼 and variable 𝑥. (a + x)1 = a x (a + x) 2 = a 2 ax x2 ( a + x )3 = a 3 a2 x ax 2 x3 (a + x) 4 = a 4 a3 x a2 x2 ax3 x4 ( a + x )5 = a 5 a4 x a3 x 2 a 2 x3 ax 4 x5 When we read the equations from left to right, we observe that (1) the powers of a decrease as the power of x increase (2) the sum of the powers of each term is the same as the power of ( a + x) 1  Topic 1: Binomial Expansion and Hayah Next, consider the coefficients: (a + x)0 = 1 (a + x) = 1 1 1 (a + x) 2 = 1 2 1 These arrangement are called (a + x) = 3 1 3 3 1 Pascal’s triangle (a + x) 4 = 1 4 6 4 1 ( a + x )5 = 1 5 10 10 5 1 Noted that (1) Each row begins and ends with 1. (2) The other coefficients in a particular row are obtained by adding the two coefficients just above in the preceding row. n We can use   to represent the coefficient for the (r + 1) term of expansion (a + x)n , where r = 0, 1, 2, r 3,... , n. n n where n! = n ( n − 1)( n − 2 )( n − 3) 1 n!   = Cr = r (n − r )!r ! Example n n Simplify (i)   = nC2 (ii)   = nCr  2 r SOLUTION n n! n  (n − 1)  (n − 2)! n  (n − 1) (i)   = nC2 = = =  2 (n − 2)!2! (n − 2)!2! 2! (ii) n n n!   = Cr = r (n − r )!r ! n  (n − 1)  (n − 2).......(n − (r − 1))  (n − r )! = (n − r )!r ! n  (n − 1)  (n − 2).......(n − r + 1) = r! 2  Topic 1: Binomial Expansion and Hayah EXAMPLE 1.1 Simplify (n − 2)! 1 (n − 3)!(n − 2)! 1 ( a) ans : (b) ans : 2 n! n(n − 1) (n!) 2 n (n − 1)2 (n − 2) n! (n − 5)! (c) ans : n(n − 1)(n − 2) (d ) ans : (n − 5)(n − 6)(n − 7) (n − 3)! (n − 8)! SOLUTION 3  Topic 1: Binomial Expansion and Hayah 1.1.1 Binomial Theorem Generally, for any positive expansion of (a + x)n , n n n n  n (a + x) =   a n x 0 +   a n −1 x1 +   a n − 2 x 2 + +   a n−r x r + n +   xn 0 1  2 r  n n n −1 n(n − 1) n − 2 2 n(n − 1)(n − 2)...(n − r + 1) n −r r = an + a x+ a x +... + a x +.. + x n 1! 2! r! In special case when a = 1, n ( n − 1) 2 n ( n − 1)( n − 2 ) 3 (1 + x ) = 1 + nx + x + x +..... + x n n 2! 3! Binomial Theorem For n a positive integer n n ( + )    a n−r x r n a x = r =0  r  Observations: 1. The expansion of (a + x)n has (n+1) terms. 2. The power of a decrease by 1 for each term as we move from left to right. 3. The power of x increase by 1 for each term as we move from left to right. 4. In each term, the sum of the powers of a and x always adds up to n.  n  n −( r −1) r −1 5. the r th term, Tr =  a x and  r − 1 n the (r+1)th termTr+1 =   a n − r x r r 4  Topic 1: Binomial Expansion and Hayah EXAMPLE 1.2 Find the expansion for the following functions (a) (2 + x)5 (b) (1 − 2 x)4 SOLUTION 5  5 5  5  5  5 (2 + x) =   25 x 0 +   24 x1 +   23 x 2 +   22 x3 +   21 x 4 +   20 x5 5 a) 0 1  2  3  4  5 = 32 + 80 x + 80 x + 40 x + 10 x + x 2 3 4 5  4  4  4  4  4 (1 − 2 x ) =  14 ( −2 x ) +  13 ( −2 x ) +  12 ( −2 x ) +  1( −2 x ) +  10 ( −2 x ) 4 0 1 2 3 4 b) 0 1  2  3  4 = 1 − 8 x + 24 x 2 − 32 x3 + 16 x 4 EXAMPLE 1.3 Find in ascending powers of x, the first four terms of the expansion the following: a) (2 + 4 x )5 b) (3 − 5 x )4 Ans : a) 32 + 320 x + 1280 x 2 + 2560 x 3 +... b) 81 − 540 x + 1350 x 2 − 1500 x 3 +... SOLUTION 5  Topic 1: Binomial Expansion and Hayah 1.1.2 Finding a particular term in a binomial expansion To obtain a particular term in a binomial expansion, we will use: n The (r+1)th term Tr+1 =   a n − r x r r EXAMPLE 1.4 Find the coefficients of x 6 and y 5 in the expansion of ( x + y ). 9 SOLUTION 9 The ( r + 1) th term =   x 9− r y r r For the term consists x 6 : 9 − r = 6  r = 3 9 Therefore, the coefficient of x 6 is   y 3 = 84y 3  3 For the term consists y 5 , r = 5 9 Therefore, the coefficient of y 5 is   x 9 −5 = 126x 4 5 EXAMPLE 1.5 6  1 Determine the middle term of  x 2 − . Ans: -20x3  x SOLUTION 6  Topic 1: Binomial Expansion and Hayah EXAMPLE 1.6 Determine the terms independent of x in the following binomial expansions 8 10 15  1  3  1  a)  x +  b)  x 2 − 2  c)  3x 3 −    x  x   2x 2  SOLUTION 8 8 8 r 1 a) The ( r + 1) term is   x8−r   =   x8− r − r =   x8− 2 r th r x r r For the term independent of x, 8 − 2r = 0  r = 4 8 Thus, the term independent of x is   = 70  4 10  3  10  r 10 − r  term is   ( x 2 )  − 2  =   ( x ) 20 − 2 r b) The ( r + 1) ( −3) x −2 r th r r  x  r 10  =  (x ) 20 − 4 r ( −3 ) r r For the term independent of x , 20 − 4r = 0  r = 5  10  Thus, the term independent of x is   ( −3 ) = 61 236 5 5 c) 7  Topic 1: Binomial Expansion and Hayah EXAMPLE 1.7 n  1 5 Given that the 4 term in the expansion of  px +  is th where n is positive integer. Find n  x 2 1 and p. Ans : n = 6 , p = 2 SOLUTION 8  Topic 1: Binomial Expansion and Hayah EXAMPLE 1.8 Find the first 4 terms of the expansion of (1 + x ) and (1 − x ). Hence find the value of 6 6 (1.001) and ( 0.999 ) in six decimal points.(6 d.p) 6 6 SOLUTION 6 6 6 (1 + x ) = 1 +   x +   x 2 +   x3 + 6 1  2  3 = 1 + 6 x + 15 x + 20 x 2 3 (1 − x ) = 1 − 6 x + 15 x 2 − 20 x3 + 6 (1.001) = (1 + 0.001) 6 6 = 1 + 6 ( 0.001) + 15 ( 0.001) + 20 ( 0.001) 2 3 = 1.006005 ( 0.999 ) = (1 − 0.001) 6 6 = 1 − 6 ( 0.001) + 15 ( 0.001) − 20 ( 0.001) 2 3 = 0.994015 9  Topic 1: Binomial Expansion and Hayah EXAMPLE 1.9 5  x Expand  2 −  in ascending power of x. Use the first four terms of your expansion to find an  2 approximation to (1.99)5 in 5 decimal points. 5 1 Ans : 32 − 40 x + 20 x 2 − 5 x3 + x 4 − x5 , 31.20796 8 32 SOLUTION 10  Topic 1: Binomial Expansion and Hayah 1.2 Binomial Theorem when n is not a positive integer LEARNING OUTCOMES At the end of the lecture, students should be able to: Expand (1 + x ) for x  1 where n is not a positive integer. n State the range of values of x for the expansion to be valid. It is known that when n is positive integer, n ( n − 1) 2 n ( n − 1)( n − 2 ) 3 (1 + x ) = 1 + nx + x + x +..... + x n n 2! 3! When is not a positive integer, the above expansion is still true, but the number of terms in the series is infinite, that is, n ( n − 1) 2 n ( n − 1)( n − 2 ) 3 (1 + x ) = 1 + nx + x + x + n 2! 3! valid for x  1 −1  x  1 EXAMPLE 1.10 Find the expansions of the following function in ascending powers of x up to and including the term in x 3 and state the range of x for which the expansion valid. 1 −2 (a) (1 + x) (b) (1 − x) 2 SOLUTION −2 ( −2 − 1) 2 −2 ( −2 − 1)( −2 − 2 ) 3 (a) (1 + x ) = 1 + ( −2 ) x + −2 x + x + 2! 3! = 1 − 2 x + 3 x 2 − 4 x 3 +.... valid for x  1 11  1  1  1  1  − 1  − 1 − 2  = [1 + (− x)] = 1 + (− x) +  2 2  (− x) 2 +  1 2 2  2  (− x)3 + (b) (1 − x ) 2 1 2 2 2! 3! 1 1 1 = 1 − x − x 2 − x3 +.... 2 8 16 valid for x  1 11  Topic 1: Binomial Expansion and Hayah EXAMPLE 1.11 Find the first three terms in the expansion of 1 (a) (8 − x) 3 (b) (2 + 3x)−1 State the range of x for which the expansion is valid. x x2 Ans : (a) 2 − − + , x 8 12 288 SOLUTION −1  3  ( 2 + 3x ) −1 −1 (b) = 2 1 + x   2  1  3  ( −1)( −1 − 2 )  3   2 = 1 + ( −1)  x  +  x +  2  2  2! 2   1 3 9   = 1 − x + x2 +  2 2 4  1 3 9 = − x + x2 + 2 4 8 3 2 valid for x  1 x 2 3 12  Topic 1: Binomial Expansion and Hayah EXAMPLE 1.12 Expand (1 + 3x ) 3 in ascending powers of x until and including the term x 3. 1 1 By substituting x = , find the value of 3 2 to 5 decimal points. 125 SOLUTION 11  1  1  1   − 1  − 1 − 2  33  3  3  3  (3 x )3 +... (1 + 3x ) = 1 + (3x) + 1 (3x )2 + 1 3 3 2! 3! 5 = 1 + x − x2 + x3 3 1 When x = , 125 1 1 1 1  3  3  128  3 64 3 2 3 4 3 (1 + 3x ) 1 = 1 +  =  = = 2 3  125   125  1 125 3 5 2 3 4 1  1  5 1  Thus, 3 2 = 1 + +  +   + 5 125  125  3  125  5  2 3 1  1  5 1  3 2 = 1 + −  +   +  4  125  125  3  125   5 = (1.007936512....) = 1.25992 to 5 decimal places. 4 13  Topic 1: Binomial Expansion and Hayah EXAMPLE 1.13 1 Expand (1 − 4 x ) in ascending powers of x up to the term in x3. 2 1 By letting x = in the series, evaluate 6 correct to four decimal places. 100 Ans: 1 − 2 x − 2 x2 − 4 x3 , 2.4495 SOLUTION 14  Topic 1: Binomial Expansion and Hayah EXAMPLE 1.14 1 − 1 1 x  2 (a) Show that = 1 −  9− x 3 9  1 −  x 2 (b) Write down the first three terms in the binomial expansion of 1 −  in ascending  9 powers of x. State the range of values of x for which the expansion is valid. x x2 Ans : 1 + + +..., where − 9  x  9 18 216 3(1 − x) (c) Find the first three terms in the expansion of in ascending powers of x, for small 9− x 17 x 11x 2 values of x. Ans : 1 − − −..., 18 216 5 (d) By letting x = 0.2 in the expansion in (c), find the value of correct to 4 decimal 11 places. Ans : 0.6742 SOLUTION 15  Topic 1: Binomial Expansion and Hayah 1.3 HAYAH Pascal’s triangle is the arrangement of numbers in a triangular form. Long before the existence of calculators, computers and phones, this arrangement is a simple yet effective method that enables the ancient civilisation to determine the coefficients in the expansion of any binomial expression, such as (𝑥 + 𝑦)𝑛. It is coined by the French mathematician Blaise Pascal, yet it is far older. This remarkable pattern of coefficients was also discovered in the 11th century by the Islamic polymath, Omar Khayyam. A polymath person is an expert in various kinds of fields. Omar Khayyam is not only an expert in mathematics but also well known as an astronomer, philosopher and poet. One of the applications of Pascal's triangle is in the arrangement of life (hayah) and death (mawt) of a certain number of people. We know for a certain that every life will experience death and return to Allah as stated in the Quran, surah Anbiya, verse 35. Every soul will taste death. And We test you with evil and with good as trial; and to Us you will be returned. (Al-Anbya: 35) Few conditions need to be met in Pascal's triangle such as every life is independent of each other. This signifies the Islamic faith that the death of person A does not cause the death of person B. As Muslims, we strongly believe that life and death as well as Qadr is in the hand of Allah alone. Abdullah (b. Mas'ud) reported that Allah's Messenger (‫ )ﷺ‬who is the most truthful (of the human beings) and his being truthful (is a fact) said: Verily your creation is on this wise. The constituents of one of you are collected for forty days in his mother's womb in the form of blood, after which it becomes a clot of blood in another period of forty days. Then it becomes a lump of flesh and forty days later Allah sends His angel to it with instructions concerning four things, so the angel writes down his livelihood, his death, his deeds, his fortune and misfortune. By Him, besides Whom there is no god, that one amongst you acts like the people deserving Paradise until between him and Paradise there remains but the distance of a cubit, when suddenly the writing of destiny overcomes him and he begins to act like the denizens of Hell and thus enters Hell, and another one acts in the way of the denizens of Hell, until there remains between him and Hell a distance of a cubit that the writing of destiny overcomes him and then he begins to act like the people of Paradise and enters Paradise. (Sahih Muslim, 2643) 16  Topic 1: Binomial Expansion and Hayah So, let’s denote P as alive and Q as death. As we know, a person can either live or dies. Then, the probability of a person dying in the next year, q can be represented as the equation (1) below. 𝑞 =1−𝑝 (1) where p is the probability of a person alive in the next year. This is only for one person. What about if we want to know about the death of four individuals? Given that, all individuals have the same probability of death and being alive. Table 1 demonstrates the application of Pascal’s Triangle in the application of life and death. Table 1 The arrangement of life and death in Pascal's Triangle Number of individuals Number of outcomes Pascal’s triangle 1 P 1 Individual A Q (A can either die or survives) 1 2 PP (A and B survive) 1, Individual A PQ, QP (only one of them 2, Individual B survives) QQ (both die) 1 3 PPP (everyone survives) 1, Individual A, B and C PPQ, PQP, QPP (two persons survive) 3, PQQ, QPQ, QQP (only one person 3, survives) 1 QQQ (everyone dies) 4 PPPP (everyone survives) 1 Indiviual A,B, C and PPPQ, PPQP, PQPP, QPPP (3 survives and 1 dies) 4 D PPQQ, QQPP, QQPP,PQPQ, QPQP, PQQP, QPPQ 6 (only 2 survive and the other 2 persons die) QQQP, QQPQ, QPQQ, PQQQ (only 1 survives) 4 QQQQ (everyone dies) 1 We could see that Table 1 corresponds with Figure 1. The first row of Table 1 matches the second row in Figure 1 and the last row in Table 1 is the same as the last row of Figure 1. From this, we can calculate the probability of death. Let's assume that there are four people in a room and the probability of death, q is equal to the probability of being alive, p. Then, 𝑝 + 𝑞 = 1 ⇒ 𝑝 = 𝑞 = 0.5. From Table 1, we have a total of 16 (1+4+6+4+1) outcomes. 4 Thus, the probability of only one person surviving is = 0.25 16 Figure 1 Pascal’s triangle for n=4 17  Topic 1: Binomial Expansion and Hayah 18