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
About The Author
Edu Delight Tutors
Am a dedicated educator with a passion for learning and a keen interest in technology. I believe that technology can revolutionize education and am committed to creating an online hub of knowledge, inspiration, and growth for both educators and students. Welcome to Edu Delight Tutors, where learning knows no boundaries.
