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.