60 Higher Engineering Mathematics
When a = 2 and n =7:
(2 + x) 7 = 2 7 + 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 + 448x + 672x 2 + 560x 3
+ 280x 4 + 84x 5 + 14x 6 + x 7
Problem 4. Use the binomial series to determine
the expansion of (2a − 3b)
5 .
From equation (1), the binomial expansion is given by:
(a + x)
n
= a
n
+ na
n−1 x +
n(n − 1)
2!
a
n−2 x
2
+
n(n − 1)(n − 2)
3!
a
n−3 x
3
+ · · ·
When a = 2a, x = −3b and n = 5:
(2a − 3b)
5
= (2a)
5
+ 5(2a)
4
(−3b)
+
(5)(4)
(2)(1)
(2a)
3
(−3b)
2
+
(5)(4)(3)
(3)(2)(1)
(2a)
2
(−3b)
3
+
(5)(4)(3)(2)
(4)(3)(2)(1)
(2a)(−3b)
4
+
(5)(4)(3)(2)(1)
(5)(4)(3)(2)(1)
(−3b)
5
i.e. (2a − 3b) 5 = 32a 5 −240a 4 b + 720a 3 b
2
−1080a 2 b
3
+ 810ab 4 −243b 5
Problem 5. Expand
c −
1
c
5
using the binomial
series.
c −
1
c
5
= c 5 + 5c 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 − 5c
3
+ 10c −
10
c
+
5
c 3 −
1
c 5
Problem 6. Without fully expanding (3 + x) 7 ,
determine the fifth term.
The r’th term of the expansion (a + x) n is given by:
n(n − 1)(n − 2)...to (r − 1) terms
(r − 1)!
a
n−(r−1) x
r−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 = 945x 4
Problem 7. Find the middle term of
2 p −
1
2q
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 the r’th term where a = 2 p,
x =−
1
2q
, n =10 and r − 1 = 5 gives:
(10)(9)(8)(7)(6)
(5)(4)(3)(2)(1)
(2 p)
10–5
−
1
2q
5
= 252(32 p
5
)
−
1
32q 5
Hence the middle term of
2 p −
1
2q
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.