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