BINOMIAL EXPANSION

THIRD TERM E-LEARNING NOTE

SUBJECT: FURTHER MATHEMATICS CLASS: SS 2

SCHEME OF WORK

 

WEEK THREE

TOPIC:BINOMIAL EXPANSION: PASCAL TRIANGLE, BINOMIAL THEOREM OF NEGATIVE, POSITIVE AND FRACTIONAL POWER

PASCAL’S TRIANGLE

Consider the expressions of each of the following:

(x + y)0; (x + y )1; (x + y)2; (x + y)3; (x + y)4

(x + y)0 = 1

(x + y)1 = 1x + 1y

(x + y)2 = 1x2 + 2xy + 1y2

(x + y)3= 1x3 + 3x2y + 3xy2 + 1y3

(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1x4

The coefficient of x and y can be displayed in an array as:

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

The array of coefficients displayed above is called Pascal’s triangle, and it is used in determining the co-efficients of the terms of the powers of a binomial expression

Coefficient of (x + y)0 1

Coefficient of (x + y)1 1 1

Coefficients of (x + y)2 1 2 1

Coefficients of (x + y)3 1 3 3 1

Coefficients of (x + y)4 1 4 6 4 1

Example 1

Using Pascal’s riangle, expand and simplify completely: (2x + 3y)4

Solution:

(2x + 3y)4 = (2x)4 + 4(2x)3 (3y) + 6(2x)2(3y)2 + 4(2x)(3y)3 + (3y)4

= 16x4 + 96x3y + 216x2y2 + 216xy3 + 81y4

Examples 2:

Using pascal’s triangle, the coefficients of (x + y)5are: 1,5,10,10,5,1.

Therefore (x – 2y)5 = x5 + 5x4(-2y) + 10x3(-2y)2 + 10x2(-2y)3 + 5x(-2y)4 + (-2y)5

= x5 – 10x4y + 40x3y2 – 80x2y3 + 80xy4 – 32y5

Example 3

Using Pascal’s triangle, simplify, correct to 5 decimal places (1.01)4

Solution

We can write (1.01)4 = (1 + 0.01)4

(1 + 0.01)4 = 1 + 4(0.01) + 6(0.01)2 + 4(0.01)3+(0.01)4

= 1 + 0.04 + 0.0006 + 0.000004 + 0.00000001

= 1.04060401

= 1.04060 (5 d.p)

The Binomial Expansion Formula

Consider the expansion of (x + y)5 again

(x + y)5 = (x + y)(x + y)(x + y)(x + y)(x + y)

The first term is obtained by multiplying the xs in the five brackets. there is only one way to doing this

(x + y)n = xn + nxn – 1y + xn – 2y2 + xn-3y3 + ….

xn-ryr + …. yn

It can be shown that the binomial expansion formula holds for positive, negative, integral or any rational value of n, provided there is a restriction on the values of x and y in the expansion of (x + y)n

We shall however consider only the binomial expansion formula for a positive integral n

Example 4:

Write down the binomial expansion of 6 simplifying all the terms

Use the expansion in (a) to evaluate (1.0025)6 correct to five significant figures.

Solution

6 = 1 + 6C1 1 + 6C22

+ 6C3 3 + 6C4

+ 6C5 5 + 6C6 6

6= 1 + x + 3 + x

4 + x 5 + 6

= 1 + x + x2+ x3 +x4 + x5 + x6

(1.0025)6 = (1 + 0.0025)6

= )6

= )6

Put x =

x = x 4 = = 0.01

therefore (1.0025)6 = 1 + (0.01) + (0.01)2 + (0.01) +(0.01)4 + …

= 1 + 0.015 + 0.00009375 + 0.0000003125

= 1.0150940625

= 1.0151 (5 s.f.)

EVALUATION

Expand ( 2 + 4x )4 simplifying the terms

Example 5
(a) Using the binomial theorem, obtain the expansion of (1 + 3x)6 + (1 – 3x)6 simplifying all the terms

(b)Use the above result to calculate the value of (1.03)6 + (0.97)6, correct to five decimal places

Solution:

(1 + 3x)6 = 1 + 6C1 (3x) + 6C2 (3x)2 + 6C3 (3x)3 _ 6C4 (3x)4 _ 6C5 (3x)5 + 6C6 (3x)6 ….. (1)

(1 – 3x)6 = 1 – 6C1 (3x) + 6C2 (3x)2 – 6C3 (3x)3 + 6C4 (3x)4 _ 6C5 (3x)5 + 6C6 (3x)6 ….. (2)

Adding (1) and (2)

(1 + 3x)6 +(1 – 3x)6 = 2 + 2 x 6C2 (3x)2 + 2 x 6C4 (3x)4 + 2 x 6C6 (3x)6

= 2 + 2 x 9x2 + 2 x x 81x4 + 2 x 729x6

= 2 + 270x2 + 2430x4 + 1458x6

(1.03)6 = (1 + 0.03)6

(0.97)6 = (1 – 0.03)6

Put 1 + 0.03 = 1 + 3x

Therefore 3x = 0.03

Therefore x = 0.01

Hence

(1.03)6 + (0.97)6 = 2 + 270(0.01)2 + 2430(0.01)4 + 1458(0.01)

= 2 + 0.027 + 0.0000243 + 2.0270243

= 2.02702 (5 d.p)

Example 6

Using the binomial theorem, expamd (1 + 2x)5, simplifying all the terms

Use your expansion to calculate the value of 1.025, correct to six significant figures

If the first three terms of the expansion of (1 + px)n in ascending powers of x are 1 + 20x + 160x,

Find the values of n and p

Solution:

(1 + 2x)5 = 1 . 5C1(2x) + 5C2(2x)2 + 5C3(2x)2 + 5C4(2x)4 + 5C5(2x)5

= 1 + 5.(2x) + . 4x2 + . 8x3 + . 16x + 32x5

= 1 + 10x + 40x2 + 80x3 + 80x4 + 32x5

(1.02) = (1 + 0.02)

Put 1 + 0.02 = 1 + 2x

Therefore 2x = 0.02

x = 0.01

Hence:

(1.02)5 = 1 + 10(0.01) + 40(0.01)2 + 80(0.01)3 + 80(0.01)4 + 32(0.01)5

= 1 + 0.1 + 0.004 + 0.0008 + 0.00000008

= 1.10408 (6.s.f.)

6.3 The Binomial Theorem for any index

The Binomial expansion formula is also applicable to any index n, where n can be positive or negative integer or even a fraction

If /x/ 1, then:

(1 + x)n = 1 + nx + + + x4 + … where n may be a negative integer or a fraction.

Example 7

Use the Binomial expansion formula to obtain the first five terms of the expansion of (1 + x)-2

Solution:

(1 + x)-2 = 1 + (-2) ( x) + ( ( x)2 + ( x)3 + ( x)4 + ….

= 1 – x + 3. 2 – 4.3 + 5.4

(1 + px)n = 1 + 20x + 160x2 + …

(1 + px)n = 1 + nc1 (px) + nc1 (px)2 + …

= 1 + npx + p2x2 …

= 1 + 20x + 160x2 + …

By equating coefficients

np = 20 … (1)

p2 = 160 … (2)

From (1) p = … (3)

Therefore p2 = … (4)

Substituting (4) into (2)

x = 160

x 200 = 160

There 200(n – 1) = 160n

200n – 200 = 160n

200n – 160n = 200

40n = 200

n = 5

From (3)p = = 4

Hence, n = 5, p = 4

Example 8

Obtain the first four terms of the explanation of (2 + x)8in ascending powers of x. hence, find the value of (2.005)8, correct to five significant figures.

Solution:

(2 + x)8= 28(1+ x)8

= 28[1 +8C1( x) + 8C2 ( )2 + 8C3 ( )3 + … ]

= 28[1 +8( x) + ( )2 + ( )3 + …]

= 28[1 + 2X + X2+ X3 + …]

Write 2.0.005

Put 2 + x = 2 + 0.005

Therefore x = 0.005

Therfore x = 0.005 x 2

= 0.01

Hence,

(2.005)8 = 28[1 +2(0.01) + (0.01)2+ (0.01)3 ]

(2.005)8 = 28 + 29(0.01) + 26.7(0.01)2 + 25 x 7(0.01)3 + …

= 256 + 5.12 + 0.0448 + 0.000224

= 261.165025

= 261.17 (5 s.f.)

GENERAL EVALUATION

1) Write down and simplify all the terms of the binomial expansion of ( 1 – x )6 . Use the expansion to evaluate 0.9976 correct to 4 dp

2) Write down the expansion of ( 1 + ¼ x ) 5 simplifying all its coefficients

3) Use the binomial theorem to expand ( 2 – ¼ x)5 and simplify all the terms

4) Deduce the expansion of ( 1 – x +x2 )6 in ascending powers of x

Reading Assignment

New Further Maths Project 2 page 73 – 78

WEEKEND ASSIGNMENT

If the first three terms of the expansion of ( 1 + px )n in ascending powers of x are 1 + 20v + 160x find the value of

1) n a) 2 b) 3 c) 4 d) 5

2) p a) 2 b) 3 c) 4 d) 5

3) In the expansion of ( 2x + 3y )4 what is the coefficient of y4 a) 16 b) 81 c) 216 d) 96

4) How many terms are in the expansion of ( 1 – 4x ) 5 a) 3 b) 5 c) 6 d) 8

5) What is the third term in the expansion of ( 1 – 3x )6 in ascending powers of x a) 18 b) -540 c) 135 d) 729

THEORY

1) Using binomial theorem, write down and simplify the first seven terms of the expansion of ( 1 + 2x )10 in ascending powers of x

2) Expand ( 2 + x )5 ( 1 – 2x ) 6 as far as the term in x3 . Evaluate ( 1.999 )5 ( 1.002 )6