Question
What is the sum of the series $$1 + \frac{1}{8} + \frac{{1.3}}{{8.16}} + \frac{{1.3.5}}{{8.16.24}} + .....\,\infty \,?$$
A.
$$\frac{2}{{\sqrt 3 }}$$
B.
$${2\sqrt 3 }$$
C.
$$\frac{{\sqrt 3 }}{2}$$
D.
$$\frac{1}{{2 \sqrt 3 }}$$
Answer :
$$\frac{2}{{\sqrt 3 }}$$
Solution :
As given the series is
$$S = 1 + \frac{1}{8} + \frac{{1.3}}{{8.16}} + \frac{{1.3.5}}{{8.16.24}} + .....\,\infty $$
On comparing this series with
$$S = {\left( {1 + x} \right)^n} = 1 + nx + \frac{{n\left( {n - 1} \right)}}{{2\,!}}{x^2} + .....\,\infty ,$$
we get,
$$\eqalign{
& nx = \frac{1}{8}\,\,\,\,\,.....\left( 1 \right) \cr
& {\text{and }}\frac{{n\left( {n - 1} \right)}}{{2\,!}}{x^2} = \frac{{1.3}}{{8.16}}\,\,\,\,\,.....\left( 2 \right) \cr} $$
From Eqs. (1) and (2), we get
$$\eqalign{
& \frac{{\frac{{n\left( {n - 1} \right)}}{{2\,!}}{x^2}}}{{{n^2}{x^2}}} = \frac{{\frac{{1.3}}{{8.16}}}}{{\frac{1}{8} \cdot \frac{1}{8}}} \cr
& \Rightarrow \frac{{n - 1}}{{2n}} = \frac{3}{2} \cr
& \Rightarrow n - 1 = 3n \cr
& \Rightarrow n = - \frac{1}{2} \cr} $$
On putting this value in Eq. (i)
$$\eqalign{
& \Rightarrow \left( { - \frac{1}{2}} \right)x = \frac{1}{8} \cr
& \Rightarrow x = - \frac{1}{4}. \cr
& {\text{But, }}S = {\left( {1 + x} \right)_n} = {\left( {1 - \frac{1}{4}} \right)^{ - \frac{1}{2}}} \cr
& = {\left( {\frac{3}{4}} \right)^{ - \frac{1}{2}}} = \frac{2}{{\sqrt 3 }}. \cr} $$