Binomial expansion

Question 1

Find the first 4 terms of the binomial expansion, in ascending powers of x, of

( 1 + x / 4 )^8

giving each term in its simplest form.

Answer 1

• a = 1
• b = x / 4
• n = 8
• ( 1 )^8 + ( 8 1 ) x ( 1 )^7 x ( x / 4 ) + ( 8 2 ) x ( 1 )^6 x ( x / 4 )^2 + ( 8 3 ) x ( 1 )^5 x ( x / 4 )^3
• = 1 + 8x / 4 + 28x^2 / 16 + 56x^3 / 64
• = 1 + 2x + 7 / 4x^2 + 7 / 8x^3

Question 2

Use your expansion to estimate the value of ( 1.025 )^8, giving your answer to 4 decimal
places.

( 1 + x / 4 )^8

( 1 + 2x + 7 / 4x^2 + 7 / 8x^3 )

Answer 2

• 1 + x / 4 = 1.025 ( The equation in the brackets is equal to the value inside the brackets )
• x / 4 = 0.025
• x = 0.1
• 1 + 2 x ( 0.1 ) + 7 / 4 x ( 0.1 )^2 + 7 / 8 x ( 0.1 )^3
  ( Substitute x value into the expanded equation )
• = 1.218375
• = 1.22

Question 3

Find the first 3 terms, in ascending powers of x, of the binomial expansion of

( 2 - 9x )^4,

giving each term in its simplest form.

Answer 3

• a = 2
• b = - 9x
• n = 4
• ( 2 )^4 + ( 4 1 ) ( 2 )^3 ( -9x ) + ( 4 2 ) ( 2 )^2 ( -9x )^2
• = 16 - 288x + 1944x^2

Question 4

f ( x ) = ( 1 + kx ) ( 2 – 9x )^4, where k is a constant.

The expansion, in ascending powers of x, of f( x ) up to and including the term in x^2 is

A - 232x + Bx^2

where A and B are constants.

b ) Write down the value of A.

c ) Find the value of k.

d ) Hence find the value of B.

Answer 4

• ( 2 - 9x )^4 = 16 - 288x + 1944x^2

( All the multiples that make x values = the x coefficient in the given equation )

• ( 1 + kx ) x ( 16 - 288x + 1944x^2 )
• • 232x = -288x + 16kx
• ( 1 x - 288x = -288x )
• ( kx x 16 = 16kx )
• 56x = 16kx
• 56 = 16k
• k = 7 / 2
• ( 1 + 7 / 2x ) x ( 16 - 288x + 1944x^2 )
• = 16 - 288x + 1944x^2 + 56x - 1008x^2
• = 16 - 232x + 936x^2
• A = 16
• K = 7 / 2
• B = 936

Question 5

( 1 - 3 / 4x )^-5 / 3 up to and including x^3 term

Answer 5

• a = 1
• b = - 3 / 4x
• n = - 5 / 3
• 1 + ( - 5 / 3 ) ( - 3 / 4x ) / 1! + ( - 5 / 3 ) ( - 8 / 3 ) ( - 3 / 4x )^2 / 2! + ( - 5 / 3 ) ( - 8 / 3 ) ( - 11 / 3 ) ( - 3 / 4x )^3 / 3!
• = 1 + 5 / 4x + 5 / 4x^2 - 55 / 48x^3
• ( General formula = a + ( n ) ( b ) / 1! + ( n ) ( n - 1 ) ( b )^2 / 2! ….. )
• ( a is only included once, n is constantly minused by 1 in coefficient and the power of b is always equal to the factorial )

Question 6

f ( x ) = 1 + x / 1 - 2x

Show that the series expansion of f ( x ) up to and including the x^3 term is 1 + 3x + 6x^2 + 12x^3

Answer 6

• f ( x ) = ( 1 + x ) x ( 1 - 2x )^-1
• a = 1
• b = - 2x
• n = - 1
• 1 + ( - 1 ) x ( - 2x ) / 1! + ( - 1 ) x ( - 2 ) x ( - 2x )^2 / 2! + ( - 1 ) x ( - 2 ) x ( - 3 ) x ( - 2x )^3 / 3!
• = ( 1 + x ) ( 1 + 2x + 4x^2 + 8x^3 )
• = 1 + 2x + 4x^2 + 8x^3 + x + 2x^2 + 4x^3 + 8x^4
• = 1 + 3x + 6x^2 + 12x^3

Question 7

State the range of values of x for which the expansion is valid

( f ( x ) = 1 + x / 1 - 2x )

Answer 7

• ( 1 - 2x )^-1
• [ - 2x ] < 1 ( x value is always < 1 )
• x < 1 / 2 ( Modules takes positive values only )

Question 8

Content on 1 + x :

Answer 8

• ( 4 + x )^1 / 2 ( Take 4 out )
• 4^1 / 2 x ( 1 + x / 4 )^1 / 2
• = 2 ( 1 + x / 4 )^1 / 2
• expand ( 1 + x / 4 )^1 / 2 using binomial, then times that expansion by “ 2 “ as it’s the coefficient of the bracket

Question 9

h ( x ) = 6 / 1 + 5x - 4 / 1 - 3x

Find the series expansion of h ( x ), in ascending powers of x, up to and including the x^2 term.
Simplify each of them.

Answer 9

• 6 / 1 + 5x - 4 / 1 - 3x
• 6 x ( 1 + 5x )^- 1 - 4 x ( 1 - 3x )^- 1
• ( 1 + 5x )^- 1
• a = 1
• b = 5x
• n = - 1
• 1 + ( - 1 ) x ( 5x ) / 1! + ( - 1 ) x ( -2 ) x ( 5x )^2 / 2!
• = 1 - 5x + 25x^2
• 6 x ( 1 - 5x + 25x^2 )
• = 6 - 30x + 150x^2
• ( 1 - 3x )^- 1
• a = 1
• b = -3x
• n = -1
• 1 + ( - 1 ) x ( - 3x ) / 1! + ( - 1 ) x ( -2 ) x ( - 3x )^2 / 2!
• = 1 + 3x + 9x^2
• = 4 x ( 1 + 3x + 9x^2 )
• = 4 + 12x + 36x^2
• ( 6 - 30x + 150x^2 ) - ( 4 + 12x + 36x^2 )
• = 2 - 42x + 114x^2

Question 10

Find the percentage error made in using the series expansion in part a to estimate the value of h ( 0.01 ).
Give your answer to 2 significant figures

• ( Expansion = 2 - 42x + 114x^2 )
• ( Original equation = 6 / 1 + 5x - 4 / 1 - 3x )

Answer 10

• 2 - 42 x ( 0.01 ) + 114 x ( 0.01 )^2 = 6 / 1 + 5 x ( 0.01 ) - 4 / 1 - 3 x ( 0.01 ) ( Equate the equations and substitute x value )
• 1.5914 = 1.590574374
• 1.5914 - 1.590574374 = 8.25626 x 10^-4
• 8.25626 x 10^-4 / 1.590574374 x 100 = 0.05190741241 % ( Answer / smaller error )
• % error = 0.052%

Question 11

Find the binomial expansion up to and inc,using the x^3 term of ( 4 + x )^- 1 / 2

Answer 11

• ( 4 + x )^- 1 / 2
• = 4^- 1 / 2 ( 1 + X / 4 )^- 1 / 2
• ( 4 is taken out of the bracket to make it in the form of ( 1 + x ) )
• ( The number outside the bracket takes the power of the bracket )
• 1 / 2 ( 1 + x / 4 )^- 1 / 2
• a = 1
• b = x / 4
• n = - 1 / 2
• 1 + ( - 1 / 2 ) ( x / 4 ) / 1! + ( - 1 / 2 ) ( - 3 / 2 ) ( x / 4 )^2 / 2! + ( - 1 / 2 ) ( - 3 / 2 ) ( - 5 / 2 ) ( x / 4 )^3 / 3!
• = 1 / 2 ( 1 - x / 8 + 3 / 128x^2 - 5 / 1024x^3 )
• = 1 / 2 - 1 / 16x + 3 / 256x^2 - 5 / 2048x^3

Question 12

State the range of values of x for which binomial expansions of the following are valid:

a ) ( 1 + 3x )^- 1 / 2

b ) ( 4 + 3x )^- 4

Answer 12

a )

• | 1 + 3x | < 0
• | 3x | < - 1
• | x | < - 1 / 3
• | x | < 1 / 3
• ( Positive x is taken )

b )

• | 4 + 3x | < 0
• | 3x | < - 4
• | x | < - 4 / 3
• | x | < 4 / 3
• ( Positive x is taken )