BINOMIAL EXPANSION

PASCAL TRIANGLE, BINOMIAL THEOREM OF NEGATIVE, POSITVE 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)

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:

a.             Write down the binomial expansion of  6 simplifying all the terms

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

(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)

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

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