The Binomial Theorem

THE BINOMIAL THEOREM

1. Binomial Expansions A binomial is an expression of the form (a+ b )n . Consider the expansion of these binomials for various values of n. (a+ b )0 = 1 1 (a+ b )1 = a+ b 1 1 (a+ b )2 = a2+2 ab + b 2 1 2 1 (a+ b )3 = a3+3 a 2 b + 3 ab 2 + b 3 1 3 3 1 (a+ b )4 = a4+4 a 3 b + 6 a 2 b 2 + 4 ab 3 + b 4 1 4 6 4 1 Notice the powers of a descend as the powers of b ascend, and the sum of the powers is always n. The triangle of numbers is known as Pascal’s triangle : the sum of each pair of adjacent numbers gives the number underneath the pair. The numbers in Pascal’s triangle correspond to the coefficients in the binomial expansions.

Example 1 : Use Pascal’s triangle to expand (2+ 3x )5 . We need the next row in the table, and this has coefficients 1, 5, 10, 10, 5, 1. 1 25 = 1 32 = 32 5 24 (3x )1 = 5创 16 3x = 240x 10 23 (3x )2 = 10创 8 9x2 = 720x2 10 22 (3x )3 = 10创 4 27x3 = 1080x3 5 21 (3x )4 = 5创 2 81x4 = 810x4 1 (3x )5 = 1 243x5 = 243x5 And so (2+ 3x )5 = 32 + 240 x + 720 x 2 + 1080 x 3 + 810 x 4 + 243 x 5 . 4 骣 1 Example 2 : Use Pascal’s triangle to expand 琪5x - . 桫 x We need the row with coefficients 1, 4, 6, 4, 1. 1 (5x )4 = 1 625x4 = 625x4 骣 1 1 4 (5x )3 琪- = 4创 125x3 - = -500x2 桫 x x 2 2 骣 1 2 1 6 (5x ) 琪- = 6创 25x 2 = 150 桫 x x 3 1 骣 1 1 20 4 (5x ) 琪- = 4创 5x - 3 = - 2 桫 x x x 4 骣 1 1 1 1 琪- = 1 4 = 4 桫 x x x 4 骣 14 2 20 1 And so 琪5x- = 625 x - 500 x + 150 - + . 桫 x x2 x 4 Example 3 : Find the coefficient of x3 in the expansion of (7- 2x )5 . Here we do not need to find the full expansion, just the term in x3 . 1 5 10 10 72 (- 2x )3 = 10创 49- 8x3 = -3920x3 5 1 The coefficient of x3 is −3920.

Example 4 : Find the coefficient of x5 in the expansion of (3x + 1)7 . We need to continue Pascal’s triangle down a few rows. 1 7 21 (3x )5 12 = 21创 243x5 1 = 5103x5 The coefficient of x5 is 5103.

Example 5 : Find the coefficient of x2 in the expansion of (1- 2x )(3 x - 5)3 . We first expand (3x - 5)3 … 1 (3x )3 = 1 27x3 = 27x3 3 (3x )2 (- 5)1 = 3创 9x2 - 5 = -135x2 3 (3x )1 (- 5)2 = 3创 3x 25 = 225x 1 (- 5)3 = 1� 125 = -125 And then multiply by (1- 2x ) , only bothering to find those terms involving x2 . 27x3 -135x2 225x -125 1 -135x2 -2x -450x The coefficient of x2 is −585. Note that if we were far-sighted enough, we would only bother finding the terms in x and x2 in the expansion of (3x - 5)3 .

C2 p72 Ex 5A

2. Formulas for Binomial Coefficients The numbers in Pascal’s triangle are called binomial coefficients. In most cases, it is not too much trouble to write the triangle out as far as you need, but for large powers – the value of n in (a+ b )n – this would be time-consuming. So we will now obtain formulas for binomial coefficients. But first, some notation. Consider the term involving br in the expansion of (a+ b )n . We use the

骣n n notation 琪 , alternatively denoted Cr in probability theory, to identify the coefficient, and so 桫r

骣n n- r r this particular term is 琪 a b . 桫r The table below shows Pascal’s triangle rewritten with n and r. r 0 1 2 3 4 5 6 0 1 1 1 1 2 1 2 1 n 3 1 3 3 1 4 1 4 6 4 1 5 1 5 10 10 5 1 6 1 6 15 20 15 6 1 骣6 骣4 骣5 So for example, 琪 =15 , 琪 =1, 琪 = 5 . 桫2 桫0 桫4 The formula for the binomial coefficients is... 骣n n! 琪 = 桫r r!( n- r) ! For example, 骣7 7! 琪 = 桫2 2!5! 7创 6 5 创 4 3 创 2 1 = 2创 1 5 创 4 3 创 2 1 = 21 n 7 All scientific calculators have an Cr button. On a graphic calculator, to get C2 , type 7 OPTN F6 F3 (to get PROB) F3 (to get nCr) 2 EXE.

C2 p73 Ex 5B We can write binomial expansions in this notation.

3骣3 3 骣 3 2 骣 3 2 骣 3 3 (a+ b ) =琪 a + 琪 a b + 琪 ab + 琪 b 桫0 桫 1 桫 2 桫 3

5骣5 5 骣 5 4 骣 5 3 2 骣 5 2 3 骣 5 4 骣 5 5 (a+ b ) =琪 a + 琪 a b + 琪 a b + 琪 a b + 琪 ab + 琪 b 桫0 桫 1 桫 2 桫 3 桫 4 桫 5 In general we have the binomial theorem...

n骣n n 骣 n n-1 骣 n n - 2 2 骣 n n - 1 骣 n n (a+ b ) =琪 a + 琪 a b + 琪 a b + ... + 琪 ab + 琪 b 桫0 桫 1 桫 2 桫n- 1 桫 n

Example 1 : Use the binomial theorem to expand (3+ 2x )5 .

5骣5 5 骣 5 4 骣 5 3 2 骣 5 2 3 骣 5 4 骣 5 5 (3+ 2x ) =琪 3 + 琪 3 (2 x ) + 琪 3 (2 x ) + 琪 3 (2 x ) + 琪 3(2 x ) + 琪 (2 x ) 桫0 桫 1 桫 2 桫 3 桫 4 桫 5 =1� 243 � 5 � 162x � 10 � 108 x2 10 72 x 3 5 48 x 4 32 x 5 =243 + 810x + 1080 x2 + 720 x 3 + 240 x 4 + 32 x 5 You can get the coefficients from your calculator or straight from Pascal’s triangle if you prefer.

4 骣 1 Example 2 : Expand 琪2x - . 桫 x2 4 2 3 4 骣1骣44 骣 4 3 骣 1 骣 4 2 骣 1 骣 4 骣 1 骣 4 骣 1 琪2x- =琪 (2 x ) + 琪 (2 x ) 琪 - + 琪 (2 x ) 琪 - + 琪 (2 x ) 琪 - + 琪琪 - 桫x2桫0 桫 1 桫 x 2 桫 2 桫 x 2 桫 3 桫 x 2 桫 4 桫 x 2 4 2 1 =1� 16x4 � 4 � 8 x � 6 4 1 x2 x 5 x 8 24 8 1 =16x4 - 32 x + - + x2 x 5 x 8

Example 3 : Find the term in x3 in the expansion of (2+ 3x )8 The required term is

骣8 5 3 3 琪 2 (3x )= 56创 32 27 x 桫3 = 48384x3

Example 4 : Find the term in x4 in the expansion of (7x - 2)9 The required term is

骣9 4 5 4 琪 (7x ) (- 2) = 126创 2401 x - 32 桫5 = -9680832x3 6 骣 1 Example 5 : Find the constant term in the expansion of 琪5x + . 桫 x2 1 You may be able to see that we need to raise 5x to the power 4 and to the power 2. x2 So the required term is

2 骣6 4骣1 4 1 琪 (5x )琪 = 15创 625 x 桫2 桫x2 x 4 = 9375

12 Example 6 : Find the term in x8 in the expansion of ( x+ x ) .

Here, we need to raise x to the power 4 and x to the power 8. So the required term is

8 骣12 4 4 4 琪 x( x) = 495创 x x 桫8 = 495x8

Also useful is a special case of the binomial theorem...

n骣n 骣 n2 骣 n 3 n (1+x ) = 1 +琪 x + 琪 x + 琪 x + ... + x 桫1 桫 2 桫 3 …since all the powers of 1 are all worth 1.

4 4 Example 7 : Use the binomial theorem to expand (1+ x ) . Hence write (1+ 3) in its simplest form. 4骣4 骣 4 2 骣 4 3 4 (1+x) = 1 +琪( x) + 琪( x) + 琪 ( x) + ( x ) 桫1 桫 2 桫 3 =1 + 4x + 6 x + 4 x x + x2 4 (1+ 3) = 1 + 4 3 + 6� 3 � 4 3 3 32 =28 + 16 3

C2 p75 Ex 5C Q1-6, p76 Ex 5D Q1 3. Approximations Using the Binomial Theorem If successive terms of a binomial expansion get smaller and smaller, we can ignore negligible terms and hence make approximations. Example 1 : Expand (1+ 2x )5 up to and including the term in x3 . Hence find an approximation for 1.025 .

5骣5 骣 5 2 骣 5 3 (1+ 2x ) = 1 +琪 (2 x ) + 琪 (2 x ) + 琪 (2 x ) + ... 桫1 桫 2 桫 3 =1 + 10x + 40 x2 + 80 x 3 + ... Substituting x = 0.01, (1.02)5 = 1 + 10� 0.01 � 40 � 0.0001 80 0.000001 ... 1.10408 This compares favourably with the true value of 1.1040808032

Example 2 : Find the expansion of (2- 3x )10 up to and including the term in x3 . Hence find an approximation for 1.9710 .

10骣10 10 骣 10 9 骣 10 8 2 骣 10 7 3 (2- 3x ) =琪 2 + 琪 2 ( - 3 x ) + 琪 2 ( - 3 x ) + 琪 2 ( - 3 x ) + ... 桫0 桫 1 桫 2 桫 3 =1024 - 15360x + 103680 x2 - 414720 x 3 + ... Substituting x = 0.01, 1.9710 = 1024 - 15360� 0.01 � 103680 � 0.0001 414720 0.000001 ... 880.35328 This agrees with the true value to one decimal place.

Example 3 : If x is so small that terms of x3 and higher can be ignored, show that (x+ 2)(3 - 2 x )5� 486 + 1377 x 1350 x 2 . Expanding the binomial, (3- 2x )5淮 1 3 5 + 5 � 3 4 ( + 2 x ) � 10 3 3 ( 2 x ) 2 �243+ 810x 1080 x2 We then multiply by (x + 2) , only bothering to find those terms involving x2 or lower. 243 -810x 1080x2 x 243x -810x2 2 486 -1620x 2160x2 The expansion is therefore 486- 1377x + 1350 x2 . The graphs on the next page show that this is a reasonable approximation for values of x close to zero. The solid line is y=( x + 2)(3 - 2 x )5 and the dotted line is y=486 - 1377 x + 1350 x2 . 512005. 24680 0 01–. 2468 01. 2468

y 2000

– 1 – 0.8 – 0.6 – 0.4 – 0.2 0.2 0.4 0.6 0.8 1 x

C2 p75 Ex 5C Q7-8, p76 Ex 5D Q2,3,5

4. Problems Where the Power is Unknown We can write... 骣n n! 琪 = = 1 桫0 0!n ! 骣n n! 琪 = = n 桫1 1!(n - 1)! 骣n n! n ( n - 1) 琪 = = 桫2 2!(n - 2)! 2! 骣n n! n ( n- 1)( n - 2) 琪 = = 桫3 3!(n - 3)! 3! ...and so on. This is useful when the power of the binomial is unknown, as in the following examples.

Example 1 : In the expansion of (1+ 3x )n , the coefficient of x2 is 105. Find the value of n.

骣n 2 2 琪 (3x )= 105 x 桫2 n( n - 1) �9 105 2! n( n - 1) = 30 n2 - n -30 = 0 (n- 6)( n + 5) = 0 n = 6 We reject the negative root (for now – we will accept it when we get to unit C4!) Example 2 : In the expansion of (1+ px )q , the coefficients of x and x2 are –28 and 336 respectively. Find the values of p and q. q( q - 1) (1+px )q = 1 + q ( px ) + ( px )2 + ... 2! Comparing coefficients, we have pq = -28 q( q - 1) p2 = 336 2! Rearranging the first equation… 28 p = - q 784 p2 = q2 …and substituting in the second, q( q - 1) 784 � 336 2! q2 392(q - 1) = 336 q 392q- 392 = 336 q 56q = 392 q = 7 p = -4

Example 3 : In the expansion of (a+ x )(1 + x )n , the first two terms are 3+ 16x . Find the values of a and n, and find the coefficients of x2 and x3 .

n 骣 n( n- 1)2 n ( n - 1)( n - 2) 3 (a+ x )(1 + x ) = ( a + x )琪 1 + nx + x + x + ... 桫 2! 3!

骣an( n- 1)2 骣 n ( n - 1) an ( n - 1)( n - 2) 3 =a +(1 + an ) x +琪 n + x + 琪 + x ... 桫2! 桫 2! 3! The first two terms are 3+ 16x , and so a = 3 1+an = 16 1+ 3n = 16 n = 5 Substituting into the formulas above for the coefficients of x2 and x3 gives us 3创 5 4 coefficient of x2 = 5 + = 35 2 5创 4 3 5 创 4 3 coefficient of x3 = + = 40 2 6

C2 p76 Ex 5D Q4, p77 Ex 5E Q1,6 Topic Review : Binomial Theorem