62 NUMBER AND ALGEBRA
= 1 +(−3)(2x)+
(−3)(−4)
2!
(2x)^2
+
(−3)(−4)(−5)
3!
(2x)^3 +···
= 1 − 6 x+ 24 x^2 − 80 x^3 +···
(b) The expansion is valid provided| 2 x|<1,
i.e. |x|<
1
2
or−
1
2
<x<
1
2
Problem 11.
(a) Expand
1
(4−x)^2
in ascending powers ofx
as far as the term inx^3 , using the binomial
theorem.
(b) What are the limits of x for which the
expansion in (a) is true?
(a)
1
(4−x)^2
=
1
[
4
(
1 −
x
4
)] 2 =
1
42
(
1 −
x
4
) 2
=
1
16
(
1 −
x
4
)− 2
Using the expansion of (1+x)n
1
(4−x)^2
=
1
16
(
1 −
x
4
)− 2
=
1
16
[
1 +(−2)
(
−
x
4
)
+
(−2)(−3)
2!
(
−
x
4
) 2
+
(−2)(−3)(−4)
3!
(
−
x
4
) 3
+···
]
=
1
16
(
1 +
x
2
+
3 x^2
16
+
x^3
16
+···
)
(b) The expansion in (a) is true provided
∣
∣
∣
x
4
∣
∣
∣<1,
i.e.|x|< 4 or− 4 <x< 4
Problem 12. Use the binomial theorem to
expand
√
4 +xin ascending powers ofxto
four terms. Give the limits ofxfor which the
expansion is valid.
√
4 +x=
√[
4
(
1 +
x
4
)]
=
√
4
√(
1 +
x
4
)
= 2
(
1 +
x
4
)^1
2
Using the expansion of (1+x)n,
2
(
1 +
x
4
)^1
2
= 2
[
1 +
(
1
2
)(
x
4
)
+
(1/2)(− 1 /2)
2!
(x
4
) 2
+
(1/2)(− 1 /2)(− 3 /2)
3!
(x
4
) 3
+···
]
= 2
(
1 +
x
8
−
x^2
128
+
x^3
1024
−···
)
= 2 +
x
4
−
x^2
64
+
x^3
512
−···
This is valid when
∣
∣
∣
x
4
∣
∣
∣<1,
i.e. |x|<4or− 4 <x< 4
Problem 13. Expand
1
√
(1− 2 t)
in ascending
powers oftas far as the term int^3.
State the limits oftfor which the expression is
valid.
1
√
(1− 2 t)
=(1− 2 t)−
1
2
= 1 +
(
−
1
2
)
(− 2 t)+
(− 1 /2)(− 3 /2)
2!
(− 2 t)^2
+
(− 1 /2)(− 3 /2)(− 5 /2)
3!
(− 2 t)^3 +···,
using the expansion for (1+x)n
= 1 +t+
3
2
t^2 +
5
2
t^3 +···
The expression is valid when| 2 t|<1,
i.e. |t|<
1
2
or −
1
2
<t<
1
2