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.

Using @RobPratt's elegant solution
$$S_1(x)=\sum_{k=1}^\infty \frac{\binom{2k}{k}}{k^2 }x^k=2 x \, _4F_3\left(1,1,1,\frac{3}{2};2,2,2;4 x\right)$$ Simplifying
$$S_1(x)=2 x \left(\frac{\text{Li}_2\left(\frac{1}{2}-\frac{1}{2} \sqrt{1-4
   x}\right)}{x}-\frac{\left(\log \left(1+\sqrt{1-4 x}\right)-\log (2)\right)^2}{2
   x}\right)$$
$$S_1(4)=2 \text{Li}_2\left(\frac{1}{2} \left(1-i \sqrt{15}\right)\right)+\left(\tan
   ^{-1}\left(\sqrt{15}\right)-i \log (2)\right)^2$$
$$\color{blue}{S_1(4)=\frac {\pi^2}6-2 \log ^2(2)}$$
$$S_2(x)=\sum_{k=1}^\infty (-1)^k \frac{\binom{2k}{k}}{k^2 }x^k=-2 x \, _4F_3\left(1,1,1,\frac{3}{2};2,2,2;-4 x\right)$$ Simplifying
$$S_2(x)=-2 x \left(\frac{\left(\log \left(1+\sqrt{1+4 x}\right)-\log (2)\right)^2}{2
   x}-\frac{\text{Li}_2\left(\frac{1}{2}-\frac{1}{2} \sqrt{1+4 x}\right)}{x}\right)$$
$$S_2(4)=2 \text{Li}_2\left(\frac{1}{2} \left(1-\sqrt{17}\right)\right)-\log ^2(2)-\log
   \left(1+\sqrt{17}\right) \text{csch}^{-1}(4)$$
$$\color{blue}{S_2(4)=2 \text{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)-\left(\sinh ^{-1}(1)-\log (2)\right)^2}$$
Combining all of the above
$$\color{red}{2(S_1(4)+S_2(4))=4 \text{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)+\frac{\pi ^2}{3}-6 \log
   ^2(2)+2 \sinh ^{-1}(1) \left(2 \log (2)-\sinh ^{-1}(1)\right)}$$
