Higher Engineering Mathematics

60 NUMBER AND ALGEBRA

Problem 4. Expand

(

c−

1

c

) 5

using the bino-

mial 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 5. 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

Substituting n=7, a=3 and r− 1 = 5 − 1 = 4

gives:

(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 6. 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 =5 gives:

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

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

(2p)10–5

(

−

1

2 q

) 5

=252(32p^5 )

(

−

1

32 q^5

)

Hence the middle term of

(

2 p−

1

2 q

) 10

is− 252

p^5

q^5

Problem 7. Evaluate (1.002)^9 using the bino-

mial theorem correct to (a) 3 decimal places and

(b) 7 significant figures.

(1+x)n= 1 +nx+

n(n−1)

2!

x^2

+

n(n−1)(n−2)

3!

x^3 +···

(1.002)^9 =(1+ 0 .002)^9

Substituting x= 0 .002 and n=9 in the general

expansion for (1+x)ngives:

(1+ 0 .002)^9 = 1 +9(0.002)+

(9)(8)

(2)(1)

(0.002)^2

+

(9)(8)(7)

(3)(2)(1)

(0.002)^3 +···

= 1 + 0. 018 + 0. 000144

+ 0. 000000672 +···

= 1. 018144672 ...

Hence (1.002)^9 =1.018, correct to 3 decimal

places

=1.018145, correct to

7 significant figures

Problem 8. Evaluate (0.97)^6 correct to 4 sig-

nificant figures using the binomial expansion.

(0.97)^6 is written as (1− 0 .03)^6

Using the expansion of (1+x)nwheren=6 and

x=− 0 .03 gives:

(1− 0 .03)^6 = 1 +6(− 0 .03)+

(6)(5)

(2)(1)

(− 0 .03)^2

+

(6)(5)(4)

(3)(2)(1)

(− 0 .03)^3

+

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

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

(− 0 .03)^4 +···

= 1 − 0. 18 + 0. 0135 − 0. 00054

+ 0. 00001215 −···

≈ 0. 83297215