A sum of Fibonacci numbers Let $F_n$ be the Fibonacci numbers. I would like to prove this really messy identity: 
$$\sum_{n=0}^{\infty}(-1)^n(n+1)^2\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n+2\choose 1}{2n+2 \choose 2}{2n \choose n}}=F_{k}\tag1$$
which simplifies to:
$$\sum_{n=0}^{\infty}(-1)^n\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n \choose n}(4n+2)}=F_{k}$$
How can I evaluate the following sums to show this identity (1) is correct?
$$\sum_{n=0}^{\infty}(-1)^n\frac{n^2}{{2n \choose n}(2n+1)}=A$$
$$\sum_{n=0}^{\infty}(-1)^n\frac{n}{{2n \choose n}(2n+1)}=B$$
$$\sum_{n=0}^{\infty}(-1)^n\frac{1}{{2n \choose n}(2n+1)}=C$$
 A: Using Wolfram for these sums, we get approximate values
\begin{align*}
C \approx 0.86082, \qquad B \approx -0.11649, \qquad A \approx -0.07897 
\end{align*}
From there, you can collect coefficients of $F_r$ to prove your identity. A more concrete way to derive exact expressions is through generating functions. Define
\begin{align*}
g(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{\binom{2n}{n}(2n+1)}x^n
\end{align*}
which you can show equals
\begin{align*}
g(x) = 4 \frac{\sinh^{-1}(\sqrt{x}/2)}{\sqrt{x(x+4)}}
\end{align*}
If we define $L = x \frac{d}{dx}$ as the operator which first takes the derivative w.r.t $x$, then multiplies by $x$, we get that
\begin{align*}
C = (L^0 g)(1), \qquad B = (L^1 g)(1), \qquad A = (L^2 g)(1)
\end{align*}
or
\begin{align*}
C = 4 \frac{\sinh^{-1}(1/2)}{\sqrt{5}}, \qquad B = \frac{2}{5} - \frac{12 \sinh^{-1}(1/2)}{5\sqrt{5}}, \qquad A = \frac{4}{125}(7\sqrt{5}\sinh^{-1}(1/2) - 10)
\end{align*}
A: For the strict purpose of the question, we do not need $A,B,C$, but some linear combinations of them,
that have "easy values". This answer goes mainly in this direction.

Some words in advance.
First of all let us write the correct form of the relation:
$$
\sum_{n=0}^{\infty}
(-1)^n(n+1)^2
\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}
{{2n+2\choose 1}{2n+2 \choose 2}{2n \choose n}}
=
\color{red}{F_{k+1}}\qquad(1)
\ .
$$
The simplified version of it is thus:
$$
\sum_{n=0}^{\infty}
(-1)^n
\frac
{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}
{{2n \choose n}\cdot 2(2n+1)}
=
\color{red}{F_{k+1}}\qquad(2)
\ .
$$
This should hold for all $k$. So let us isolate in the denominator of $(2)$
an expression involving only two Fibonacci numbers, my choice being $F_k$ and $F_{k+1}$.
We split for this $5n^2 = (5n^2-1)+\color{blue}1$ and form thus from $\color{blue}{1}\cdot F_k+F_{k-1}$ one $F_{k+1}$.
The other Fibonacci numbers can the be easily linearly written in terms of $F_k$ and $F_{k+1}$ (using small Fibonacci numbers
as coefficiente), e.g. $F_{k+3}=2F_{k+1}+F_k$, so we have to show:
$$
\sum_{n\ge0}
(-1)^n
\frac
{(5n^2+4n+1)F_k + (5n+3)F_{k+1}}
{{2n \choose n}\cdot2(2n+1)}
=
\color{red}{F_{k+1}}\qquad(3)
\ .
$$
We have thus to show that the coefficients of $F_k,F_{k+1}$ extracted from the expression in the L.H.S. are $0,2$, as in the R.H.S.
We completely avoid hypergeometric sums for this, so let us finally start the answer.

We use the notations
$$
a_n= \frac 1{\binom{2n}n}\ ,\qquad
b_n= \frac n{\binom{2n}n}\ ,
$$
and compute simply
$$
\begin{aligned}
a_n+a_{n+1}
&=
\frac 1{\binom{2n}n}
\left[1+\frac 1{\ \frac{(2n+1)(2n+2)}{n+1)(n+1)}\ }\right]
=
\frac {5n+3}{\binom{2n}n \cdot2(2n+1)}
\ ,
\\
b_n+b_{n+1}
&=
\frac 1{\binom{2n}n}
\left[n+\frac {n+1}{\ \frac{(2n+1)(2n+2)}{n+1)(n+1)}\ }\right]
=
\frac {5n^2+4n+1}{\binom{2n}n \cdot2(2n+1)}
\ ,
\end{aligned}
$$
and no the first relation 
already realizes the telescope for the coefficient of $F_{k+1}$, for short this is
$\sum_{n\ge 0}(-1)^n(a_n+a_{n+1}) = a_0=1$.
We simimilarly use the second relation to see that the coefficient of $F_k$ is $b_0=0$. Done.
$\square$

If the question insists to have the values for $A,B,C$, then let us some lines.
It is enough to get the value of $A$.
sage computes for it the exact value and the approximation:
sage: var('n');
sage: sum( (-1)^n / binomial(2*n, n) / (2*n+1), n, 0, oo )
4/5*sqrt(5)*log(1/2*sqrt(5) + 1/2)

sage: S = sum( (-1)^n / binomial(2*n, n)          , n, 0, oo ).n()
sage: C = sum( (-1)^n / binomial(2*n, n) / (2*n+1), n, 0, oo ).n()

sage: S
0.627836423614399
sage: C
0.860817881928008
sage: 5*S + C
4.00000000000000

and we need to know something about hypergeometric series.
A: Here is another answer, in the same spirit as that of @dan_fulea:
Both sides of $$\sum_{n=0}^{\infty}(-1)^n\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n \choose n}(4n+2)}=F_{k+1}$$ can be considered as sequences defined by the same second order linear recurrence for the index $k$. Then, for them to be equal for all $k$, it suffices that they coincide at the first values $0$ and $1$ of the index $k$.
In other words, we just need to show that 
\begin{align}\sum_{n=0}^{\infty}\frac{(-1)^n}{2}\frac{5n+3}{2n+1}\frac{1}{{2n \choose n}}&= F_1=1
\end{align} 
and
\begin{align}
\sum_{n=0}^{\infty}\frac{(-1)^n}{2}\frac{5n^2+9n+4}{2n+1}\frac{1}{{2n \choose n}}=\sum_{n=0}^{\infty}\frac{(-1)^n}{2}\frac{5n+4}{2n+1}\frac{1}{C_n} &=F_2=1
\end{align}
where we have introduced the Catalan number $C_n=\frac{1}{n+1} \binom{2n}{n}$.
But it is easy to verify, by induction, that 
\begin{align}\sum_{n=0}^{m}\frac{(-1)^n}{2}\frac{5n+3}{2n+1}\frac{1}{{2n \choose n}}&=\frac{{2m+2 \choose m+1}+(-1)^m}{{2m+2 \choose m+1}}\\
\sum_{n=0}^{m}\frac{(-1)^n}{2}\frac{5n+4}{2n+1}\frac{1}{C_n}&=\frac{C_{m+1}+(-1)^m}{C_{m+1}}
\end{align} 
and we are done.
