Question

If $S_{\infty }=1+\frac{4}{3}+\frac{9}{3^2}+\frac{16}{3^3}+\cdots \cdots$, then $2S_{\infty }$ is

Answer 1

Let $S_{\infty }=1+\frac{4}{3}+\frac{9}{3^2}+\frac{16}{3^3}+\dots \dots \infty \cdots \left(i\right)$
$\frac{1}{3}{S}_{\infty }=\frac{1}{3}+\frac{4}{3^2}+\frac{9}{3^3}+\frac{16}{3^4}+\dots \dots \infty \cdots \left(ii\right)$
From $\left(i\right)-\left(ii\right)$
$\frac{2}{3}{S}_{\infty }=1+\frac{3}{3}+\frac{5}{3^2}+\frac{7}{3^3}+\dots \dots \infty \cdots \left(iii\right)\frac{2}{3^2}{S}_{\infty }=\frac{1}{3}+\frac{3}{3^2}+\frac{5}{3^3}+\dots \dots \infty \cdots \left(iv\right)$
from $\left(iii\right)-\left(iv\right)$
$\frac{4}{9}{S}_{\infty }=1+\frac{2}{3}+\frac{2}{3^2}+\frac{2}{3^3}+\dots \dots \infty =1+2\left(\frac{1/3}{1-1/3}\right)=22S_{\infty }=9$