If the value of (1+2/3+6/3^2+10/3^3+… … upto ∞)^log_(0.25)(1/3+1/3^2+1/3^3+… … upto ∞) is l, then l^2 is equal to \ \ \ \.

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