They look similar (2)

Well, what struck me was the Taylor series of
[-ln(1-x)]/x
Hence, on integrating the function at x = 1/2 gives us the answer....!!

Actually, if you see closely, the expression is simply the Dilogarithm of 0.5.........!!!!

We know $\displaystyle1+\frac x2+\frac {x^2}3+\frac{x^3}4+...=\dfrac{-\ln(1-x)}x$

Integrate it and get $\displaystyle x+\frac{x^2}4+\frac{x^3}9+\frac{x^4}{16}+...=\displaystyle\int \frac{-\ln(1-x)}x dx=f(x)$ , and $f(0)=0$

So, $\displaystyle f(x)=\int_0^x \frac{-\ln(1-n)}n dn$ , and we have to find $f(0.5)$

First, we have $f(1)=\dfrac{\pi^2}6=f(0.5)+(f(1)-f(0.5))$

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The derivative of $\ln(x)\ln(1-x)$ is $\dfrac{-\ln(x)}{1-x}-\dfrac{-\ln(1-x)}x$

So, $\displaystyle\left. _{\color{#FFFFFF}0}^{\color{#FFFFFF}0}(\ln(x)\ln(1-x))\right|_0^{0.5}=\int_{0}^{0.5}\frac{-\ln(x)}{1-x}dx-\int_{0}^{0.5}\frac{-\ln(1-x)}xdx$

$=\displaystyle\int_{0.5}^{1}\frac{-\ln(1-x)}{x}dx-\int_{0}^{0.5}\frac{-\ln(1-x)}xdx=(\ln(0.5))^2=(f(1)-f(0.5))-f(0.5)$

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Finally, $\dfrac{\pi^2}6=f(0.5)+(f(1)-f(0.5))$ , $(\ln(0.5))^2=(f(1)-f(0.5))-f(0.5)$

We have $\dfrac{\pi^2}6-(\ln(0.5))^2=2f(0.5)$ , $f(0.5)=\dfrac{\pi^2}{12}-\dfrac{(\ln(2))^2}2$

Relevant wiki: Polylogarithm

The polylogarithm function is defined as $\displaystyle \text{Li}_s (x) = \sum_{n=1}^\infty \dfrac {x^n}{n^s}$ for all complex $s$ and $|x| \le 1$ . Putting $s=2$ and $x=\dfrac 12$ , $\implies \displaystyle \sum_{n=1}^\infty \frac 1{n^2 2^n} = \text{Li}_2 \left(\frac 12 \right) = \dfrac {\pi^2}{12} - \dfrac {(\ln 2)^2}2$ . Therefore, $A+B+C+D+E+F+G = 1+2+12+1+2+2+2 = \boxed{22}$ .