Csdrhrt

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

It can be shown $|x|<4impliessum_{n=1}^inftyfrac{x^n}{n^2binom{2n}{n}}=2arcsin^2frac{sqrt{x}}{2}$, so the choice $x=phi^2$ gives $sum_{n=1}^inftyfrac{phi^{2n}}{n^2binom{2n}{n}}=2arcsin^2frac{phi}{2}=frac{9pi}{50}$ as desired. (The arcsine's evaluation is possible by chasing isosceles triangles in a regular pentagon with inscribed diagonals.)

Consider the well-known series expansion of the squared $arcsin$ function. Namely

$$2arcsin^2left(frac x2right)=sum_{n=1}^{infty}frac{x^{2n}}{n^2binom{2n}n}$$

As you can see the RHS equals your given sum for the case that $x=phi$. Plugging this in yields to

$$begin{align}
2arcsin^2left(frac phi2right)=sum_{n=1}^{infty}frac{phi^{2n}}{n^2binom{2n}n}=frac{9pi^2}{50}
end{align}$$

which further reduces the problem to

$$sinleft(frac{3pi}{10}right)=fracphi2=frac{1+sqrt{5}}{4}$$

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

$$begin{align}
sinleft(frac{3pi}{10}right)&=cosleft(frac{pi}5right)\
sinleft(frac{3pi}{10}right)&=cosleft(frac{7pi}{10}-frac{pi}2right)\
sinleft(frac{3pi}{10}right)&=sinleft(frac{7pi}{10}right)\
sinleft(frac{3pi}{10}right)&=sinleft(-frac{3pi}{10}+piright)\
sinleft(frac{3pi}{10}right)&=-sinleft(-frac{3pi}{10}right)
end{align}$$

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

$$begin{align}
x^4+x^3+x^2+x+1&=0\
x^2+frac1{x^2}+x+frac1x+1&=0\
left(x+frac1xright)^2+left(x+frac1xright)-1&=0
end{align}$$

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.

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.