Fibonacci sum for $\pi$: $\sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}}=\frac{4\pi^2}{25\sqrt5}$ Wikipedia's "List of formulae involving $\pi$" entry states that
$$\sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}}=\frac{4\pi^2}{25\sqrt5}.$$
Why is this true?
If $\varphi=(1+\sqrt5)/2$ and $\psi=(1-\sqrt5)/2$, then we can rewrite this identity as
$$F(\varphi)-F(\psi)=\frac{4\pi^2}{25}\quad\text{ where }F(x)=\sum_{n=1}^\infty\frac{x^{2n}}{n^2\binom{2n}{n}}.$$
$F(x)$ looks similar to the power series for arcsin.
 A: $$F(x)=\sum_{n=1}^\infty\frac{x^{2n}}{n^2\binom{2n}{n}}=2 \left[\sin ^{-1}\left(\frac{x}{2}\right)\right]^2$$ makes
$$F(\varphi)=\frac{9 \pi ^2}{50}\qquad \text{and}\qquad F(\psi)=\frac{\pi ^2}{50}$$
since 
$$\varphi=1-2 \cos \left(\frac{3 \pi }{5}\right)$$
A: In this answer, it is shown that
$$
\frac{\arcsin(x)}{\sqrt{1-x^2}}
=\sum_{k=0}^\infty\frac1{(2k+1)}\frac{4^k}{\binom{2k}{k}}x^{2k+1}\tag1
$$
Integrating $(1)$ and multiplying by $4$ yields
$$
\begin{align}
2\arcsin^2(x)
&=\sum_{k=0}^\infty\frac1{(2k+1)(2k+2)}\frac{4^{k+1}}{\binom{2k}{k}}x^{2k+2}\\
&=\sum_{k=0}^\infty\frac{4^{k+1}x^{2k+2}}{(k+1)^2\binom{2k+2}{k+1}}\\
&=\sum_{k=1}^\infty\frac{4^kx^{2k}}{k^2\binom{2k}{k}}\tag2
\end{align}
$$
Applying $F_n=\frac{\phi^n-(1-\phi)^n}{\sqrt5}$, we get
$$
\begin{align}
\sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}}
&=\frac2{\sqrt5}\left(\arcsin^2\left(\frac\phi2\right)-\arcsin^2\left(\frac{1-\phi}2\right)\right)\\
&=\frac2{\sqrt5}\left(\frac{9\pi^2}{100}-\frac{\pi^2}{100}\right)\\
&=\frac{4\pi^2}{25\sqrt5}\tag3
\end{align}
$$

Addendum: Trigonometric Values Used Above
Note that
$$
\begin{align}
4\cos^2(2\pi/5)+2\cos(2\pi/5)-1
&=\left(e^{2\pi i/5}+e^{-2\pi i/5}\right)^2+\left(e^{2\pi i/5}+e^{-2\pi i/5}\right)-1\\[6pt]
&=e^{4\pi i/5}+e^{2\pi i/5}+1+e^{-2\pi i/5}+e^{-4\pi i/5}\\
&=\frac{e^{10\pi i/5}-1}{e^{2\pi i/5}-1}\,e^{-4\pi i/5}\\
&=0\tag4
\end{align}
$$
Solving for the positive root of $(4)$ gives
$$
\begin{align}
\cos(2\pi/5)
&=\frac{-1+\sqrt{5}}4\\
&=\frac{\phi-1}2\tag5
\end{align}
$$
Since $\cos(2\pi/5)=2\cos^2(\pi/5)-1$ and $\cos(\pi/5)$ is positive, apply $\phi^2=\phi+1$ to get
$$
\cos(\pi/5)=\frac\phi2\tag6
$$
Since $\cos(x)=\sin(\pi/2-x)$, $(5)$ and $(6)$ become
$$
\begin{align}
\sin(\pi/10)&=\frac{\phi-1}2\tag7\\
\sin(3\pi/10)&=\frac\phi2\tag8
\end{align}
$$
