Question

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

Answer 1

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

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

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

$-$ $-$ $-$ $-$ $-$

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

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

$\frac{2S}{3}=\frac{4}{3}+\frac{4}{3^2}\left(\frac{1}{1-\frac{1}{3}}\right)$ [sum of infinite G.P]

$=\frac{4}{3}+\frac{4}{9}\left(\frac{3}{2}\right)=\frac{4}{3}+\frac{2}{3}=2$

$\Rightarrow S=3$