Higher Engineering Mathematics, Sixth Edition

60 Higher Engineering Mathematics

Whena=2andn=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

Problem 4. Use the binomial series to determine

the expansion of( 2 a− 3 b)^5.

From equation (1), 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 a,x=− 3 bandn=5:

( 2 a− 3 b)^5 =( 2 a)^5 + 5 ( 2 a)^4 (− 3 b)

+

( 5 )( 4 )

( 2 )( 1 )

( 2 a)^3 (− 3 b)^2

+

( 5 )( 4 )( 3 )

( 3 )( 2 )( 1 )

( 2 a)^2 (− 3 b)^3

+

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

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

( 2 a)(− 3 b)^4

+

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

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

(− 3 b)^5

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

− 1080 a^2 b^3 + 810 ab^4 − 243 b^5

Problem 5. Expand

(

c−

1

c

) 5

using the binomial

series.

(

c−

1

c

) 5

=c^5 + 5 c^4

(

−

1

c

)

+

( 5 )( 4 )

( 2 )( 1 )

c^3

(

−

1

c

) 2

+

( 5 )( 4 )( 3 )

( 3 )( 2 )( 1 )

c^2

(

−

1

c

) 3

+

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

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

c

(

−

1

c

) 4

+

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

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

(

−

1

c

) 5

i.e.

(

c−

1

c

) 5

=c^5 − 5 c^3 + 10 c−

10

c

+

5

c^3

−

1

c^5

Problem 6. Without fully expanding( 3 +x)^7 ,

determine the fifth term.

Ther’th term of the expansion(a+x)nis given by:

n(n− 1 )(n− 2 )...to(r− 1 )terms

(r− 1 )!

an−(r−^1 )xr−^1

Substitutingn=7,a=3andr− 1 = 5 − 1 =4gives:

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

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

( 3 )^7 −^4 x^4

i.e. the fifth term of( 3 +x)^7 = 35 ( 3 )^3 x^4 = 945 x^4

Problem 7. Find the middle term of

(

2 p−

1

2 q

) 10

.

In the expansion of(a+x)^10 there are 10+1, i.e. 11

terms. Hence the middle term is the sixth. Using the

general expression for ther’th term wherea= 2 p,

x=−

1

2 q

,n=10 andr− 1 =5gives:

( 10 )( 9 )( 8 )( 7 )( 6 )

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

( 2 p)10–5

(

−

1

2 q

) 5

= 252 ( 32 p^5 )

(

−

1

32 q^5

)

Hence the middle term of

(

2 p−

1

2 q

) 10

is− 252

p^5

q^5

Problem 8. Evaluate( 1. 002 )^9 using the binomial

theorem correct to (a) 3 decimal places and (b) 7

significant figures.