Question

If the value of ${(1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\dots \text{upto}\infty )}^{{\log }_{\left(0.25\right)}(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\dots \text{upto}\infty )}$ is $l$, then $l^2$ is equal to

Answer 1

$S_{\infty }=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\dots \infty \dots \left(1\right)$
$\frac{1}{3}{S}_{\infty }=\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+\dots \infty \dots \left(2\right)$
From $\left(2\right)-\left(1\right)$
$\frac{2}{3}{S}_{\infty }=1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\dots \infty \Rightarrow \frac{2}{3}{S}_{\infty }=\frac{4}{3}(1+\frac{1}{3}+\frac{1}{3^2}+\dots \infty )\Rightarrow \frac{2}{3}{S}_{\infty }=\frac{4}{3}\times \frac{1}{1-\frac{1}{3}}\Rightarrow S_{\infty }=3$

$\therefore$ Given expression: $l=3^{{\log }_{\left(1/4\right)}\left(1/2\right)}$
$\Rightarrow l=3^{\left(\frac{1}{2}{\log }_{\left(1/2\right)}\left(1/2\right)\right)}$
$\Rightarrow l=\sqrt{3}$
$\Rightarrow l^2=3$