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. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is B $3$

Let $S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+...$ ...(1)

Multiplying (1) by $\frac{1}{3}$, we get

$\frac{1}{3}S=\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+...$ ...(2)

$\left(1\right)-\left(2\right)$, gives

$\Rightarrow (1-\frac{1}{3})S=1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+...$
$\Rightarrow \frac{2}{3}S=1+\frac{1}{3}+\frac{4}{3^2}(1+\frac{1}{3}+\frac{1}{3^2}+...)$
$\Rightarrow \frac{2}{3}S=\frac{4}{3}+\frac{4}{9}\left(\frac{1}{1-\frac{1}{3}}\right)$ [Sum of GP$=\frac{a(1-r^n)}{1-r}$ when $r<1$]

$\Rightarrow S=3.$