Victoria School AM #6 Binomial Theorem 2024

Secondary 4 Additional Mathematics Binomial Theorem 1 For a positive integer n, ! ( 1) ( 2) ... 3 2 1n n n n=  −  −     Teacher Name: Class Reg No. Date: 4 ___ Chapter 6: Binomial Theorem Reference Book: Additional Maths 360 Textbook, Marshall Cavendish You will learn how to, • Use the notations !n and ,n r   • Expand (1 + b) n and (a + b) n to any number of terms, where n is a positive integer, • Identify and find a particular term in the expansion of (1 + b) n using the result Tr + 1 = rn      b r , • Use the result Tr + 1 = rn      a n – r b r to find a particular term in the expansion of (a + b) n , • Use the general term formula to find the specific terms, coefficients and unknown values. The factorial n! The notation n! is read as “ n factorial”. It is defined for a positive integer n as the product of the first n positive integers. In mathematical terms, Recap… Recall that ( ) 2 22 2a b a ab b+ = + + . Hence, when 1a = , we have ( ) 2 21 1 2b b b+ = + + . In pairs, expand the following and arrange the terms in ascending powers of b. (a) ( ) 3 1 b+ ( )( ) ( )( ) 2 2 2 2 3 23 1 1 1 1 2 1 2 2 1 3 3 b b b b b b b b b b b b b + + = + + + = + + + + + = + + + (b) ( ) 4 1 b+ ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( ) ( ) 4 2 2 22 43 23 43 23 2 3 2 3 2 3 2 3 4 2 3 4 1 1 1 1 2 1 2 or 1 1 1 1 1 3 3 1 1 1 1 1 3 3 1 1 3 3 1 3 3 1 3 3 3 3 1 4 6 4 b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b + = + + = + + + + + = + + = + + + + + = + + = + + + + = + + + + + + + = + + + + + + + = + + + +

Secondary 4 Additional Mathematics Binomial Theorem 2 (c) ( ) 5 1 b+ ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 5 2 3 2 2 3 2 3 2 3 2 2 3 2 3 2 3 4 2 3 4 5 2 3 4 5 1 1 1 1 2 1 3 3 1 1 3 3 2 1 3 3 1 3 3 1 3 3 2 6 6 2 3 3 1 5 10 10 5 b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b + = + + = + + + + + = + + + + + + + + + + + = + + + + + + + + + + + = + + + + + (d) ( ) 6 1 b+ ( ) ( )( ) ( )( ) ( ) ( ) 65 2 3 4 5 2 3 4 5 2 3 4 5 2 3 4 5 2 3 4 5 6 2 3 4 5 6 1 1 1 1 1 5 10 10 5 1 1 5 10 10 5 1 5 10 10 5 1 5 10 10 5 5 10 10 5 1 6 15 20 15 6 b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b + = + + = + + + + + + = + + + + + + + + + + + = + + + + + + + + + + + = + + + + + + Is there a faster way to expand ( )1 n b+ for all positive integer n?

Secondary 4 Additional Mathematics Binomial Theorem 3 Pascal Triangle Complete the following array by adding up the adjacent numbers in the preceding row above. This is known as the Pascal Triangle, named after the mathematician, Blaise Pascal, though the ancient Chinese were known to use it before he was born. One of its many interesting properties is that each row in the array corresponds to the coefficients in a binomial expansion. These coefficients are also known as binomial coefficients. The Pascal Triangle still does not give us a quick way to expand ( )1 n b+ as it takes time to construct the triangle, especially for large values of n. Therefore, we introduce the notation, n r   for the Pascal Triangle . n r   is used to find the number of ways of choosing r items from a group of n different items. When n and r are both non-negative, with the exception of 0 0   , n r   can also be written as Cn r . The first few rows of the Pascal Triangle looks like: Note that 0C1n = and C1n n = . 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1

Secondary 4 Additional Mathematics Binomial Theorem 4 Evaluating n r   ( )( ) ( )( ) 6 4 C ( 1)( 2)...( 1) ! ( 1)( 2)...( 1) 1 1 ... 2 1 1 2 2 1 ... 1 2 2 1 6 5 4 3e.g. C 15 4 3 2 1 n r n r n n n n r r n n n n r r r r n n n n r n r r r r  = − − − += − − − += −− − − − + − +=     −−   ==    (a) ( ) ( ) ( ) ( )( ) 7 2 !C , where ! 1 2 ... 2 1.!! Is C C ? Y / N 7! 7 6 5 4 3 2 1e.g. C 21 2!5! 2 1 5 4 3 2 1 n r nn r n r n r r r rr n r − = =  −  −   − =      = = =      (b) Note: Your calculator is able to evaluate Cn r using the nCr button. For example, to evaluate 6 4C , press 6 nCr 4 to get 15. Example 1: Evaluate 5 3   without the use of calculator. 5 5 4 3 3 3 2 1 10  =  = !r is read as “r factorial” and 0! is defined as 1.

Secondary 4 Additional Mathematics Binomial Theorem 5 Example 2: Given that 1203 n = . Find the value of n. ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) 2 32 32 32 3 2 2 1203 12 1203 2 1 3 2 120 6 3 2 720 0 Let f 3 2 720. f 10 10 3 10 2 10 720 0 By Factor Theorem, 10 is a factor of f . 3 2 720 10 72 By comparing the coefficie n n n n n n n n n n n n n n nn n n n n n pn  = −− = − + = − + − = = − + − = − + − = − − + − = − + + ( )( ) ( )( ) ( ) 2 2 2 nts of : 2 72 10 10 70 7 10 7 72 0 10 or 7 72 0 7 7 4 1 72 21 7 239 (NA)2 10 n p p p n n n n n n n n n =− = = − + + = = + + = −  −= −  −= = Example 3: Expand ( ) 7 1 x+ and arrange the terms in ascending powers of x. (refer to Pascal’s Triangle) ( ) 7 2 3 4 5 6 71 1 7 21 35 35 21 7x x x x x x x x+ = + + + + + + +

Secondary 4 Additional Mathematics Binomial Theorem 6 1 .r r nTb r + =  Expansion of expressions of the form ( )+1 n b for a positive integer n Notice that in the expansion of ( )1 n b+ , which is ( ) 0 1 2 3 1 1 2 3 1 1 ... ... 0 1 2 3 1 1 ... ...1 2 3 1 n r n n r n n n n n n n n nb b b b b b b b r n n n n n n nb b b b b b rn − −              + = + + + + + + + +              −                       = + + + + + + + +          −          , each term consists of two components (or factors), i.e. • the n r   component and • the rb component. Hence if we can figure out what each component of a term is, we could find that term without having to expand the entire expression. In general, for a positive integer n the term containing rb is rn br   . The term containing rb is also the ( ) th 1r + term in the expansion. i.e. Question: How many terms are there in the binomial expansion of ( )1 n b+ ? Ans: n + 1 Example 4: Expand ( ) 3 12 x+ . ( ) ( ) ( ) ( ) ( ) ( ) 3 2 3 23 23 331 2 1 2 2 2 12 1 3 2 3 4 8 1 6 12 8 x x x x x x x x x x    + = + + +       = + + + = + + +

Secondary 4 Additional Mathematics Binomial Theorem 7 Example 5: Without expanding ( ) 8 1 3 ,x− state (i) the term containing 4,x (ii) the coefficient of 7.x ( ) ( ) ( ) ( ) 1 4 4 5 4 5 4 7 7 8 7 8 7 88(i) 3 3 8When 4, 3 4 5670 The required term is 5670 . 8(ii) When 7, 3 7 17 496 The coefficient of is 17 496. rr r rT x x rr r T x Tx x r T x Tx x +    = − = −       = = −  = = = −  =− − Class Practice 1: 1 Use the Pascal’s Triangle to expand ( ) 4 1 2 .x− ( ) ( ) ( ) ( ) ( ) 4 2 3 4 2 3 4 1 2 1 4 2 6 2 4 2 2 1 8 24 32 16 x x x x x x x x x − = + − + − + − + − = − + − + 2 Use the Pascal’s Triangle to find the first four terms, in ascending powers of x, in the expansion of ( ) 9 1 2 .x+ ( ) ( ) ( ) ( ) 9 2 3 23 1 2 1 9 2 36 2 84 2 ... 1 18 144 672 ... x x x x x x x + = + + + + = + + + + 3 Use the Binomial Theorem to find the first four terms, in ascending powers of x, in the expansion of ( ) 721 2 .x− ( ) ( ) ( ) ( ) ( ) ( ) ( ) 7 2 32 2 2 2 2 4 6 2 4 6 7 7 71 2 1 2 2 2 ... 1 2 3 1 7 2 21 4 35 8 ... 1 14 84 280 ... x x x x x x x x x x      − = + − + − + − +        