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

Let S be the sum of the given series i.e., 

$\begin{array}{rl} & S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\dots .\text{to}\mathrm{\infty }\\ \Rightarrow & (S-1)=\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\dots \text{to}\mathrm{\infty }----\left(\mathrm{i}\right)\\ \Rightarrow & (S-1)\times \frac{1}{3}=\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+\dots \text{to}\mathrm{\infty }-----\left(\mathrm{ii}\right)\end{array}$

Subtracting (ii) from (i), we get 

$\frac{2}{3}(\mathrm{S}-1)=\frac{2}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\dots \text{to }\mathrm{\infty }$

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