Question

The sum to infinity of the series $1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+....$ is

• A. $3$
• B. $4$
• C. $6$
• D. $2$

Answer 1

The correct option is A $3$

Let $S=1+\frac{2}{3}+\frac{6}{3^2}+..........\left(1\right)$

$\frac{S}{3}=\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+..........\left(2\right)$

Subtracting equation $\left(2\right)$ from $\left(1\right)$, we have

$S-\frac{S}{3}=(1+\frac{2}{3}+\frac{6}{3^2}+.....)-(\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+.....)$

$\frac{2S}{3}=1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^3}+.....$

$S=\frac{3}{2}+\frac{1}{2}+\frac{2}{3}+\frac{2}{3^2}+\frac{2}{3^3}+.....$

$S=2+\frac{2}{3}+\frac{2}{3^2}+\frac{2}{3^3}+.....$

Since the above series is in G.P. of infinite terms with common ration $\frac{1}{3}$.

$\therefore S=\frac{2}{1-\frac{1}{3}}\Rightarrow S=\frac{2\times 3}{2}=3$

Hence, this is the answer.