<s>[INST] Prove that:$\sum_{n=0}^{\infty}{2^{n+3}(n^2+n+\phi)\over (n+1)(2n+1)(2n+3){2n\choose n}}=\phi\pi^2+8\phi^2\pi-8\phi^3\sqrt{5}$ $\phi$ is the golden ratio
$$\sum_{n=0}^{\infty}{2^{n+3}(n^2+n+\phi)\over (n+1)(2n+1)(2n+3){2n\choose n}}=\phi\pi^2+8\phi^2\pi-8\phi^3\sqrt{5}$$
I try:
$$S=\sum_{n=0}^{\infty}{2^{n+3}\over {2n\choose n}}\left[{A\over n+1}+{B\over 2n+1}+{C\over 2n+3}\right]$$
$n^2+n+\phi=A(2n+1)(2n+3)+B(n+1)(2n+3)+C(n+1)(2n+1)$
$A=-\phi$, $B=\phi-0.25$ and $C=\phi+0.75$
$$S=\sum_{n=0}^{\infty}{2^{n+3}\over {2n\choose n}}\left[{-\phi\over n+1}+{\phi-0.25\over 2n+1}+{\phi+0.75\over 2n+3}\right]$$
If we know the closed form
$$\sum_{n=0}^{\infty}{2^n\over {2n\choose n}(an+b)}=F(a,b)$$
Then it is easy to prove the above series.
I manage to find 
$$\sum_{n=0}^{\infty}{2^n\over {2n\choose n}(2n+1)}={\pi\over 2}$$
Any help? Thank you.
 [/INST] Too long for a comment.
$$F(a,b)=\sum_{n=0}^{\infty}{2^n\over {2n\choose n}(an+b)}=\frac 1b \,\, _3F_2\left(1,1,\frac{b}{a};\frac{1}{2},\frac{b}{a}+1;\frac{1}{2}\right)$$ where appears the generalized hypergeometric function.
$$F(2,1)=\frac \pi 2$$ $$F(2,3)=\frac{3 \pi }{2}-4$$ $$F(1,1)=\pi -\frac{\pi ^2}{8}$$
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