Question

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

Answer 1

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

$\frac{S}{3}=\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+\frac{14}{3^5}+...$ by multiplying the above equation by $\frac{1}{3}$

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

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

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

$\Rightarrow \frac{2S}{3}=\frac{4}{3}+\frac{4}{3^2}\left(\frac{1}{1-\frac{1}{3}}\right)$ using the formula $S_{\infty }=\frac{a}{1-r}$ when the terms are in G.P.

$\Rightarrow \frac{2S}{3}=\frac{4}{3}+\frac{4}{3^2}\left(\frac{3}{2}\right)$

$\Rightarrow \frac{2S}{3}=\frac{4}{3}+\frac{2}{3}=\frac{6}{3}=2$

$\Rightarrow S=\frac{3}{2}\times 2=3$

$\therefore S=3$