◙ EP-Program

3

- Strisuksa School - Roi-et

Math : Binomial Theorem

► Dr.Wattana Toutip - Department of Mathematics – Khon Kaen University © 2010 :Wattana Toutip

◙ [email protected]

◙ http://home.kku.ac.th/wattou

3 The binomial theorem 3.1 Positive integer powers Pascal’s Triangle is the triangle of numbers shown below, in which each term is obtained by adding the two terms above it. 1 n0 1 1 n 1 1 2 1 n2 1 3 3 1 n3

1 1 1

4 5

6

6 10

15

4 10

20

n4 n5 n6

1 5

15

1 6

1

The binomial coefficient nCr is the r th term of the n th row of Pascal’s Triangle, (counting from 0 instead of from 1 ). 5 4 So for example : C0  1; C2  6 ;

C6  1

6

These coefficients are also given by the formula: n! n Cr  r !(n  r )! n The binomial coefficients are sometimes written as   r The binomial theorem state that: (a  b)n  a n  nCr a n1b   nCr a nr br 

 bn .

So for example (a  b)5  a5  5a 4b  10a 3b 2  10a 2b3  5ab 4  b5 . The binomial theorem can be used to find approximations for expressions of the form (1  x)n , where x is small. Continue taking terms until they are so small that they do not affect the answer to the required degree of accuracy. 3.1.1 Examples 1. Expand ( x  2 y )4 . Solution Use the binomial theorem, with the fourth row of Pascal’s Triangle. Put a  x and b  2 y .

( x  2 y) 4  x 4  4 x3 (2 y)  6 x 2 (2 y) 2  4 x(2 y)3  (2 y) 4 (1  2 y) 4  x 4  8 x3 y  24 x 2 y 2  32 xy 3  16 y 4 .

1. Use the binomial theorem to find an approximation for (0.997)7 ,taking the first three term of expansion of (1  x)7 . Solution The first three terms of (1  x)7 are : 1  7C1 x  7C2 x2 . Put x  0.003 to obtain 1  7  0.003  21 0.0032 (0.977)7  0.979189 . 3.1.2 Exercises 1. Expand the following using the binomial theorem , and simplify as far as possible. (a) (2  b)5 (b) (1  3b)4 (c) (2 x  3 y ) 4

(d) (1  2 z )6

(e) (2a  b)7

(f) (1  0.5 x)7

2. In the following expansions, write down the appropriate terms. Simplify your answer as far as possible . (a) (1  x)8 x 3 tern (b) (1  y )9 y 5 term (c) (2  x)10 x 8 term

(d) (1  2 x)8 x 3 term

(e) (3  2 y )5 y 3 term

(f) (3x  5)5 x 2 term

(g) ( x  y )16 x5 y11 term

(h) (2a  3b)7 a5b 2 term

1 (i) ( x  )8 x 4 term x

2 (j) ( y  )7 y 3 term y

1 1 (k) ( x 2  )7 x8 term (l) ( y  2 )9 constant term. y x 3. Find the greatest coefficient of a power of x in the following expansion : (a) (1  x)17 (b) (1  2 x)7 (c) (1  3 x)8

(d) (2  3x)9

4. Expand and simplify (1  3)6  (1  3)6 ,without using a calculator. 5. Expand (1  x)12 up to the term in x 3 .Hence find approximations for : (a) 1.0212 (b) 0.9612 6. Use the binomial expansion up to the x 3 term to find approximations for : (a) 1.018 (b) 0.957 (c) 2.019 (d) 2.965 7. The first two terms of the expansion of (2  ax)n are 1024  15360x .Find a and n . b 8. The constant term in the expansion of (ax  )10 is 8064 .Find a in terms of b . x 2 9. There is no x term the expansion of (1  ax)6  (2  bx)7 .Find the ratio of a to b .



10. Use the expansion of (1  y )10 to find 1   x  x 



2 10

, up to the term in x 2 .

11. Expand the following, up to the term indicated: (a) (1  x  x 2 )7 x 2 term (b) (1  x 2 )8 x 6 term (c) (1  2 x  3x 2 )6 x 2 term

(d) (2  x  2 x 2 )7 x 2 term

(e) (1  2 x)5 (1  x) x 2 term

(f) (1  2 x)5 (1  x) x 2 term

(g) (1  x)8 (1  x 2 ) x3 term

(h) (1  x)10 (1  x  x 2 ) x3 term

3.2.Binomial theorem for other powers If n th is not a positive integer, the binomial theorem is still true , with certain restrictions .Provided that 1  x  1 :

n(n  1) x3 n(n  1)(n  2) x3   2! 3! In general, this series continues to infinity .So if only the first few terms are take then the result is only an approximation. 3.2.1 Examples (1  x)n  1  nx 

1. Find an expansion for 1  x , up to the term in x 2 .Hence find an approximation for 17 . Solution Write 1  x as (1  x)0.5 .Putting n  0.5 in the binomial expansion: 0.5(0.5  1) 2 x 2!  1  0.5 x  0.125 x 2

(1  x)0.5  1  0.5 x 

Write 17 as (16  1)0.5  4(1  0.0625)0.5 Put x  0.0625 in the expansion above :

17  4[1  (0.5)(0.0625)  (0.125)(0.0625)2 ]

 4.1230 . 1  3x 1. Expand up to the terms in x 2 . For which values of x is the expansion valid? 1 2x Solution 1  3x Write as (1  3x)(1  2 x) 1 1 2x 1x  2 x(2 x) 2 (1  2 x) 1  1  1x(2 x)  2! 2  1  2x  4x 1  3x  (1  3x)(1  2 x  4 x 2 ) . 1  2x Expand this ignoring the term in x 3 : 1  3x  1  5 x  10 x 2 1 2x The expansion is valid if 1  2 x  1 .

The expansion is valid for 0.5  x  0.5 3.2.2 Exercises 1. Expand the following up to the term x 2 . 1

(b) (1  3 x)

(a) (1  x) 4 (c) (1  x )

3 2

(d) (1  2 x)

4 3

1 1  2x 1 3 (g) (h) 2 1 x 1  x2 2. For which values of x are the expansions in Question 1 valid? 3. Use the expansion up to x 2 to find approximations for:

(e) (1  x)2

(f)

1 2

(a) 1.01

(c) 0.98

(b) 1.05

1 2

(d) 0.96

4. Without using a calculator or table , use the binomial expansion up to x 3 to find an approximation for

1.01  0.99 . 1 2

5. Using the binomial expansion up to x of (1  x) , and writing 50 as 49  1 , find an 3

approximation for 50 . 6. Use the method of Question 5 to find approximations for : (a) 101

(b) 98

1 3

(c) 28 (d) 63 2 7. Expand the following up to term in x :

1 3

1  x2 (b) 1 2x

1 x (a) 1 x 1  2 x  x2 (c) 1  3x

(d)

8. Show that the expansion of

1 x 1  x2

(1  x)  (1  x) up to the tern in x 3 is of the form a  bx 2

.Find a and b . 9.

1  ax  bx 2 has no tern in either x or x 2 .Find a and b . (1  x) 2

10. Find the expansion of

3

1 (1  x) , up to the x 2 tern. By putting x  ,show that 8

800 . 253 3.3.Examination questions 1. Find the first three terms in the expansion in ascending powers of x , of (i) (1  3x)5 (ii) (2  x) 4

approximately

3

7 is

Hence find the coefficient of x 2 in the expansion of (1  3x)5 (2  x) 4 . 2. (i) Write down the expansion of (1  x)5 .Deduce that the first three terms in the expansion of (1  3x)(1  x)5 are 1  8 x  25 x 2 and find the term in x 3 . 1 by taking the first three terms 100 of the expansion. Show that the inclusion of t he term in x 3 does not affect the approximation when working to 4 decimal places. 3. Obtain the expansion in ascending powers of x of (1  x)8 .Hence calculate the value of

(ii) Find an approximation to (1  3x)(1  x)5 when x 

(0.95)8 correct to three decimal places.

When a ball is dropped on to level ground it always rebounds to a height of 0.95 times the height from, which it fell. If it is dropped from a height of 10 metres, calculate, correct to three significant figures, the height to which it rebounds after eight bounces. 4. (a) Evaluate the tern which is independent of x in the expansion of 1 ( x  )8 x (b) Expand (1  x)2 (1  2 x) in ascending powers of x as far the term in x 3 . For which values of x is this expansion valid?

1  px 5. The first three terms in ascending power of and (1  x)10 are identical. Find the 1  qx values of p and q , assuming that x is sufficiently small for both expansions to be valid. 1

Us e this result to that the tenth root of 33 is approximately

651 2 . 649

1 in partial fractions .Hence or otherwise, given that x is (1  x)(2  x) sufficiently small for powers of x above the second to be neglected, show that 1 1  (4  6 x  7 x 2 ) (1  x)(2  x) 8 (b) Show that the first three non-zero terns in the series expansion of

6. (a) Express

1

 1 n 1    n in ascending powers of

1 are : n 2

1 11 1       n 2 n 4

3

1 and find the term in   . n By giving n a suitable value, use the first four non-zero terms of the series to find a value for 1

(0.9)10

giving your answer to five decimal places. 1 4

7. Use the binomial series to expand (1  x ) in ascending powers of x as far as the term in x3 .

1 in the expression and in the series, calculate the value of 51/4 correct to 81 five decimal places, explaining why your working ensures this degree of accuracy. Common errors 1. Positive integer powers (a) When writing out the expansion of (a  b) n do nor forget the binomial coefficients.

By putting x 

(b) When expanding something like (a  2b)5 , do not forget to take powers of 2 as of b . For example , the third term is 10a3 (2b)2  40a3b 2  20a3b 2 . (c) Similarly, be careful of negative numbers. If you are expanding (a  2b)5 , then the terms will be alternately positive and negative. 2. Other powers If n is not a positive integer, then the binomial theorem only holds for expansions of the form (1  x) n , where 1  x  1 .Do not try to expand

5 as (1  4)1/2 ,using x  4 . 1/2

Similarly, the first term inside the bracket must be 1 .Do not write

 1 1 5 as  4    2 2

and attempt to expand . 1 5 is to write it as (4  1)1/2  2(1  )1/2 . This can now be 4 expand using the Binomial theorem.

The correct way to find

Solution (Exercise) 3.1.2 1. (a) 32  80b  80b2  40b3  10b4  b5 (b) 1  12b  54b2  108b3  82b4 (c) 16 x 4  96 x3 y  216 x 2 y 2  216 xy 3  81y 4 (d) 1  12 z  60 z 2  160 z 3  240 z 4  192 z 5  64 z 6 (e) 128a 7  448a 6b  672a5b2  560a 4b3  280a3b4  84a 2b5  14ab6  b7

1 5 7 1 7 (f) 128  244 x  168 x 2  70 x3  17 x 4  2 x5  x 6  x 2 8 32 128

2. (a) 56x 3 (b) 126 y 5 (c) 180x8 (d) 448x 3 (e) 720 y 3 (f) 11250x 2 (g) 4368x5 y11 (h) 6048a5b2 (i) 28x 4 (j) 672 y 3 (k) 21x8 (l) 84 3. (a) 24310 (b) 672 (c) 20412 (d) 489888 4. 416 5. 1  12 x  66 x 2  220 x3 (a) 1.26816 (b) 0.61152 6. (a) 1.083 (b) 0.698 (c) 535.5 (d) 227.2 7. a  3, n  10 8. 9.

2 b 224 : 5

10. 1  10 x  55 x 2 11. (a) 1  7 x  28 x 2 (b) 1  8 x 2  28 x 4  56 x 6 (c) 1  12 x  78 x 2 (d) 128  488 x  1568 x 2 (e) 1  8 x  28 x 2

(f) 1  9 x  30 x 2 (g) 1  8 x  29 x 2  64 x3 (h) 1  11x  56x 2  175x3 3.2.2 1. (a) 1 

x 3x 2  4 32

3x 9 x 2  2 8 3x 3x 2 (c) 1   2 8 (b) 1 

8x 8x2  3 9 1  2 x  3x 2 1  2 x  4 x2 1  x2 3  3x 2

(d) 1  (e) (f) (g) (h)

2. a, c, e, g , h, 1  x  1 3. (a) 1.00498 (b) 1.02468 (c) 0.98995 (d) 0.9798 4. 1.999975 5. 7.07106 6. (a) 10.0499 (b) 9.8995 (c) 3.0366 (d) 3.97906 7. (a) 1  2 x  2 x 2 (b) 1  2 x  5 x 2 (c) 1  x  2 x 2 (d) 1  x  x 2 1 8. a  2, b   4 9. a  2, b  1

x x2 10. 1   3 9

1 1 b,   x  3 3

1 1 d, f ,   x  2 2

x x2 11. 1   2 8

=========================================================== References: Solomon, R.C. (1997), A Level: Mathematics (4th Edition) , Great Britain, Hillman Printers(Frome) Ltd.