Higher Engineering Mathematics

THE BINOMIAL SERIES 59

A

Using Pascal’s triangle method:

(a+x)^5 =a^5 + 5 a^4 x+ 10 a^3 x^2 + 10 a^2 x^3 +···

Hence

(2p− 3 q)^5 =(2p)^5 +5(2p)^4 (− 3 q)

+10(2p)^3 (− 3 q)^2

+10(2p)^2 (− 3 q)^3

+5(2p)(− 3 q)^4 +(− 3 q)^5

i.e.(2p− 3 q)^5 = 32 p^5 − 240 p^4 q+ 720 p^3 q^2

− 1080 p^2 q^3 + 810 pq^4 − 243 q^5

Now try the following exercise.

Exercise 32 Further problems on Pascal’s

triangle

1. Use Pascal’s triangle to expand (x−y)^7
   [
   x^7 − 7 x^6 y+ 21 x^5 y^2 − 35 x^4 y^3
   
   
   • 35 x^3 y^4 − 21 x^2 y^5 + 7 xy^6 −y^7
   
   
   
   

]

2. Expand (2a+ 3 b)^5 using Pascal’s triangle
   [
   32 a^5 + 240 a^4 b+ 720 a^3 b^2
   
   
   • 1080 a^2 b^3 + 810 ab^4 + 243 b^5
   
   
   
   

]

7.2 The binomial series

The binomial seriesorbinomial theorem is a
formula for raising a binomial expression to any
power without lengthy multiplication. The general
binomial expansion of (a+x)nis given by:

(a+x)n=an+nan−^1 x+

n(n− 1 )

2!

an−^2 x^2

+

n(n− 1 )(n− 2 )

3!

an−^3 x^3

+ ···

where 3! denotes 3× 2 ×1 and is termed ‘factorial 3’.
With the binomial theoremnmay be a fraction, a
decimal fraction or a positive or negative integer.
Whennis a positive integer, the series is finite, i.e.,
it comes to an end; whennis a negative integer, or a
fraction, the series is infinite.
In the general expansion of (a+x)nit is noted that

the 4th term is:

n(n−1)(n−2)

3!

an−^3 x^3. The number

3 is very evident in this expression.

For any term in a binomial expansion, say the

r’th term, (r−1) is very evident. It may therefore

be reasoned thatther’th term of the expansion

(a+x)nis:

n(n−1)(n−2)...to (r−1) terms

(r−1)!

an−(r−^1 )xr−^1

Ifa=1 in the binomial expansion of (a+x)nthen:

(1+x)n= 1 +nx+

n(n−1)

2!

x^2

+

n(n−1)(n−2)

3!

x^3 +···

which is valid for− 1 <x<1.

Whenxis small compared with 1 then:

(1+x)n≈ 1 +nx

7.3 Worked problems on the binomial

series

Problem 3. Use the binomial series to deter-

mine the expansion of (2+x)^7.

The binomial expansion is given by:

(a+x)n=an+nan−^1 x+

n(n−1)

2!

an−^2 x^2

+

n(n−1)(n−2)

3!

an−^3 x^3 +···

Whena=2 andn=7:

(2+x)^7 = 27 +7(2)^6 x+

(7)(6)

(2)(1)

(2)^5 x^2

+

(7)(6)(5)

(3)(2)(1)

(2)^4 x^3 +

(7)(6)(5)(4)

(4)(3)(2)(1)

(2)^3 x^4

+

(7)(6)(5)(4)(3)

(5)(4)(3)(2)(1)

(2)^2 x^5

+

(7)(6)(5)(4)(3)(2)

(6)(5)(4)(3)(2)(1)

(2)x^6

+

(7)(6)(5)(4)(3)(2)(1)

(7)(6)(5)(4)(3)(2)(1)

x^7

i.e.(2+x)^7 = 128 + 448 x+ 672 x^2 + 560 x^3

+ 280 x^4 + 84 x^5 + 14 x^6 +x^7