<s>[INST] Evaluating $\sum \limits_{k=1}^{\infty} \frac{\binom{4k}{2k}}{k^2 16^k}$ I want to find the closed form of:
$\displaystyle \tag*{}\sum \limits_{k=1}^{\infty} \frac{\binom{4k}{2k}}{k^2 16^k}$
I tried to use the taylor expansion of $\frac{1}{\sqrt{1-x}}$ and $\frac{1}{\sqrt{1-4x}}$ but both of them had $\binom{2n}{n}$ in the numerator and no square in the denominator. I am unsuccesful even in using $\arcsin^2x$ expansion.
However, we know a well known result:
$\displaystyle \tag*{} \sum \limits_{k=1}^{\infty} \frac{\binom{2n}{n}}{k^24^k} = \frac{\pi^2}{6} - 2\ln^2(2)$
So this somehows tells (?) maybe we can decompose our sum into two parts with one being Basel sum. Any help would be appreciated, thanks.
 [/INST] Note that $$\frac{1+(-1)^k}{2} = \begin{cases}1 &\text{if $k$ is even}\\ 0 &\text{if $k$ is odd}\end{cases}$$
implies that $$\sum_k a_{2k} = \sum_k \frac{1+(-1)^k}{2} a_k.$$  Taking $a_k = \frac{\binom{2k}{k}}{(k/2)^2 16^{k/2}}$ yields
\begin{align}
\sum_{k=1}^\infty \frac{\binom{4k}{2k}}{k^2 16^k}
&= \sum_{k=1}^\infty \frac{1+(-1)^k}{2}\frac{\binom{2k}{k}}{(k/2)^2 16^{k/2}} \\
&= 2\sum_{k=1}^\infty (1+(-1)^k)\frac{\binom{2k}{k}}{k^2 4^k} \\
&= 2\sum_{k=1}^\infty \frac{\binom{2k}{k}}{k^2 4^k} + 2\sum_{k=1}^\infty (-1)^k\frac{\binom{2k}{k}}{k^2 4^k} \\
\end{align}
</s>