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

$l={\left(1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\dots .. \mathrm{upto} \infty \right)}^{{\log }_{\left(0.25\right)}\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\dots . \mathrm{upto} \infty \right)}$

Assume, $S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\dots .. \mathrm{upto} \infty$

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

$\frac{2S}{3}=\frac{4}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\dots .$

$S=\frac{3}{2}\left(\frac{{\displaystyle \frac{4}{3}}}{1-{\displaystyle \frac{1}{3}}}\right)=3$

Now $l={\left(3\right)}^{{\log }_{0.25}\left(\frac{{\displaystyle \frac{1}{3}}}{1-{\displaystyle \frac{1}{3}}}\right)}$

$l=3^{{\log }_{\left(\frac{1}{4}\right)}\left(\frac{1}{2}\right)}=3^{\frac{1}{2}}=\sqrt{3}$

$\Rightarrow l^2=3$