Question
The sum to infinite term of the series $$1 + \frac{2}{3} + \frac{6}{{{3^2}}} + \frac{{10}}{{{3^3}}} + \frac{{14}}{{{3^4}}} + .....{\text{ is}}$$
A.
3
B.
4
C.
6
D.
2
Answer :
3
Solution :
$$\eqalign{
& {\text{We have}} \cr
& S = 1 + \frac{2}{3} + \frac{6}{{{3^2}}} + \frac{{10}}{{{3^3}}} + \frac{{14}}{{{3^4}}} + .....\infty \,\,\,\,\,\,......\left( 1 \right) \cr
& {\text{Multiplying both sides by }}\frac{1}{3}{\text{ we get}} \cr
& \frac{1}{3}S = \frac{1}{3} + \frac{2}{{{3^2}}} + \frac{6}{{{3^3}}} + \frac{{10}}{{{3^4}}} + .....\infty \,\,\,\,\,\,\,\,\,\,......\left( 2 \right) \cr
& {\text{Subtracting eqn}}{\text{.}}\left( 2 \right){\text{from eqn}}{\text{.}}\left( 1 \right){\text{we get}} \cr
& \frac{2}{3}S = 1 + \frac{1}{3} + \frac{4}{{{3^2}}} + \frac{4}{{{3^3}}} + \frac{4}{{{3^4}}} + .....\infty \cr
& \Rightarrow \,\,\frac{2}{3}S = \frac{4}{3} + \frac{4}{{{3^2}}} + \frac{4}{{{3^3}}} + \frac{4}{{{3^4}}} + .....\infty \cr
& \Rightarrow \,\,\frac{2}{3}S = \frac{{\frac{4}{3}}}{{1 - \frac{1}{3}}} = \frac{4}{3} \times \frac{3}{2} \cr
& \Rightarrow \,\,S = 3 \cr} $$