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

Consider

$$S=1+\frac23+\frac6{3^2}+\frac{10}{3^3}+\cdots$$

The numerators are

$$1,2,6,10,\ldots$$

whose general term is

$$4n-2
\quad\text{for }n\ge1$$

Hence,

$$S=1+\sum_{n=1}^{\infty}\frac{4n-2}{3^n}$$

$$=1+4\sum_{n=1}^{\infty}\frac{n}{3^n}-2\sum_{n=1}^{\infty}\frac1{3^n}$$

Now,

$$\sum_{n=1}^{\infty}nr^n=\frac{r}{(1-r)^2},
\qquad |r|<1$$

Taking

$$r=\frac13,$$

$$\sum_{n=1}^{\infty}\frac{n}{3^n}=\frac{\frac13}{\left(1-\frac13\right)^2}$$

$$=\frac{\frac13}{\frac49}$$

$$=\frac34$$

Also,

$$\sum_{n=1}^{\infty}\frac1{3^n}=\frac{\frac13}{1-\frac13}$$

$$=\frac12$$

Therefore,

$$S=1+4\left(\frac34\right)-2\left(\frac12\right)$$

$$=1+3-1$$

$$=3$$

Now consider

$$T=\frac13+\frac1{3^2}+\frac1{3^3}+\cdots$$

This is a G.P. with

$$a=\frac13,\qquad r=\frac13$$

Hence,

$$T=\frac{\frac13}{1-\frac13}$$

$$=\frac12$$

Therefore,

$$l=3^{\log_{0.25}\left(\frac12\right)}$$

Now,

$$\log_{0.25}\left(\frac12\right)=\log_{\frac14}\left(\frac12\right)$$

Let this equal

$$k$$

Then,

$$\left(\frac14\right)^k=\frac12$$

$$2^{-2k}=2^{-1}$$

$$k=\frac12$$

Hence,

$$l=3^{1/2}$$

Therefore,

$$l^2=3$$

Hence, the required answer is

$$\boxed3$$