If the value of $\left(1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\ldots . \text { upto } \infty\right)^{\log _{1008}\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^2}+\ldots \text { uptoss }\right)}$ is $l$, then $l^2$ is equal to $\_\_\_\_$ .
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