How to show that $\sum_{n=1}^{\infty}\frac{\phi^{2n}}{n^2{2n \choose n}}=\frac{9}{50}\pi^2$ Given:$$\sum_{n=1}^{\infty}\frac{\phi^{2n}}{n^2{2n \choose n}}=\frac{9}{50}\pi^2$$
Where $\phi=\frac{\sqrt{5}+1}{2}$
How can I we show that the above sum is correct?
I have checked numerically, it seem correct, but i don't how to proves it 
 A: It can be shown $|x|<4\implies\sum_{n=1}^\infty\frac{x^n}{n^2\binom{2n}{n}}=2\arcsin^2\frac{\sqrt{x}}{2}$, so the choice $x=\phi^2$ gives $\sum_{n=1}^\infty\frac{\phi^{2n}}{n^2\binom{2n}{n}}=2\arcsin^2\frac{\phi}{2}=\frac{9\pi}{50}$ as desired. (The arcsine's evaluation is possible by chasing isosceles triangles in a regular pentagon with inscribed diagonals.)
A: Consider the well-known series expansion of the squared $\arcsin$ function. Namely

$$2\arcsin^2\left(\frac x2\right)=\sum_{n=1}^{\infty}\frac{x^{2n}}{n^2\binom{2n}n}$$

As you can see the RHS equals your given sum for the case that $x=\phi$. Plugging this in yields to
$$\begin{align}
2\arcsin^2\left(\frac \phi2\right)=\sum_{n=1}^{\infty}\frac{\phi^{2n}}{n^2\binom{2n}n}=\frac{9\pi^2}{50}
\end{align}$$
which further reduces the problem to
$$\sin\left(\frac{3\pi}{10}\right)=\frac\phi2=\frac{1+\sqrt{5}}{4}$$
The latter one can be shown by only using the fact that $\cos(\pi/5)=1/4(1+\sqrt 5)$. It is sufficient to show that
$$\begin{align}
\sin\left(\frac{3\pi}{10}\right)&=\cos\left(\frac{\pi}5\right)\\
\sin\left(\frac{3\pi}{10}\right)&=\cos\left(\frac{7\pi}{10}-\frac{\pi}2\right)\\
\sin\left(\frac{3\pi}{10}\right)&=\sin\left(\frac{7\pi}{10}\right)\\
\sin\left(\frac{3\pi}{10}\right)&=\sin\left(-\frac{3\pi}{10}+\pi\right)\\
\sin\left(\frac{3\pi}{10}\right)&=-\sin\left(-\frac{3\pi}{10}\right)
\end{align}$$
and the last line turn out to be true since the sine function is odd.

The assumed fact concerning the value of $\cos(\pi/5)$ can be easily verified by cosidering the fifths roots of unity. To be precise consider
$$\begin{align}
x^4+x^3+x^2+x+1&=0\\
x^2+\frac1{x^2}+x+\frac1x+1&=0\\
\left(x+\frac1x\right)^2+\left(x+\frac1x\right)-1&=0
\end{align}$$
The latter can be solved by the substitution $z=1/x+x$ which boils the problem down to the solving of two quadratic equations. The real part of the first solution $($the one with positive real and imaginary part$)$ equals the cosine of $\pi/5$ and we are done.
A: Replace $\phi$ with $x$. That gives a function of $x$ with its Taylor series.
Now to find a differential equation that the Taylor series obeys.
Hint:  
On one hand, $d^2f/dx^2=\sum (2n+2)(2n+1)a_{n+1}x^{2n}$. 
On the other hand $xdf/dx=\sum 2n a_n x^{2n}$.  
Good luck with that.  As a second hint, other people have supplied the answer.
