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., 

$S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\dots$to $\mathrm{\infty }$

$\Rightarrow (S-1)=\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\dots$to $\mathrm{\infty }$       (i)

$\Rightarrow (S-1)\times \frac{1}{3}=\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+\dots$to $\mathrm{\infty }$       (ii)

Subtracting (ii) from (i), We get 

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

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