Closed form: show that $\sum_{n=1}^\infty\frac{a_n}{(n+1)4^n}=\zeta(2) $

Let $a_n$ be the sequence defined via$$ a_n=\sum_{k=1}^n{2n \choose {n-k}}\frac{1+(-1)^{k+1}}{k^2}$$then prove that$$\sum_{n=1}^\infty\frac{a_n}{(n+1)4^n}=\frac{\pi^2}{6} $$

I tried simplifying $a_n$ first, but in the end the best I got is an expression according to Wolfram, using Hypergeometric functions:$$a_n={2n\choose {n-1}}\left(_4F_3\left({{1,1,1,1-n}\atop{2,2,2+n}};1\right)+\ _4F_3\left({{1,1,1,1-n}\atop{2,2,2+n}};-1\right)\right) $$I didn't manage to simplify it further. This work does not seem helpful, since Hypergeometric functions $_4F_3$ are no easier to work with.

At this point I'm stuck. I believe representing $a_n$ with an integral would simplify the problem, but didn't manage to achieve it for now.