Proof that $\sum\limits_{k=1}^{n}(n+1-k)^2F_k^2=F_{n+1}(F_n+F_{n+2})-n-1$ $1,1,2,3,5,8,...$ for $n=1,2,3,4,...$ It is n-th Fibonacci.
Prove this sum
$$\sum_{k=1}^{n}(n+1-k)^2F_k^2=F_{n+1}(F_n+F_{n+2})-n-1$$
I try:
We know that $\sum_\limits{k=1}^{n}F_n^2=F_nF_{n+1}$
$$\sum_{k=1}^{n}(n^2+1+k^2+2n-2nk-2k)F_k^2=\sum_{k=1}^{n}F_n^2+F_{n+1}F_{n+2}-n-1$$
$$(n^2+2n)F_nF_{n+1}+\sum_{k=1}^{n}[k^2-k(2n+1)]F_k^2=F_{n+1}F_{n+2}-n-1$$
I am not sure what to do next, please any help be nice!

As noted in a comment, this sequence is the convolution of the two sequences $a_n=F_n^2$ and $b_n=(n+1)^2$, so we can multiply the two (standard) generating functions.
First, the generating function for $a_n$ — that is, $\displaystyle\sum_{n=0}^\infty F_n^2 x^n$ — is $\dfrac{x(1-x)}{(1+x)(1-3x+x^2)}$ (see, for instance, this math.SE answer ).  Likewise, the generating function for $b_n$ is $\displaystyle\sum_{n=0}^\infty (n+1)^2x^n$ $=\dfrac{(x+1)}{(1-x)^3}$ (start with $\displaystyle\sum_{i=0}^\infty x^i=\frac1{1-x}$ and then differentiate, multiply by $x$, and differentiate again).
Now, just multiply the series together: $\displaystyle\left(\sum_{n=0}^\infty a_n\right)\left(\sum_{n=0}^\infty b_n\right)$ $\displaystyle=\sum_{n=0}^\infty\left(\sum_{i=0}^na_ib_{n-i}\right)x^n$, so
$$\begin{align}
\sum_{n=0}^\infty\left(\sum_{i=0}^n F_i^2(n+1-i)^2\right)x^n &=\left(\sum_{n=0}^\infty F_n^2\right)\left(\sum_{n=0}^\infty (n+1)^2\right) \\
&= \left(\frac{x(1-x)}{(1+x)(1-3x+x^2)}\right)\left(\frac{(x+1)}{(1-x)^3}\right) \\
&= \frac{x}{(1-3x+x^2)(1-x)^2}
\end{align}$$
(Note that the sum here goes from $i=0$ rather than $i=1$, but since $F_0=0$ it doesn't make any difference.)
Now, $F_nF_{n+1} = \sum_{i=0}^nF_i^2$ (easy to prove: $F_nF_{n+1}$ $= F_n(F_n+F_{n-1})$ $=F_n^2+F_{n-1}F_n$; now just induct), so $\displaystyle\sum_{n=0}^\infty F_nF_{n+1}x^n = \frac{1}{1-x}\sum_{n=0}^\infty F_n^2x^n$ (this is another easy convolution) $=\dfrac{x}{(1+x)(1-3x+x^2)}$. Likewise, we get the generating function $\displaystyle\sum_{n=0}^\infty F_{n+1}F_{n+2} x^n$ by shifting indices here (i.e. dividing by $x$), so $$\begin{align}
\sum_{n=0}^\infty(F_nF_{n+1}+F_{n+1}F_{n+2})x^n &=\dfrac{x}{(1+x)(1-3x+x^2)} + \dfrac{1}{(1+x)(1-3x+x^2)}\\
&=\frac{x+1}{(1+x)(1-3x+x^2)} \\
&=\frac{1}{(1-3x+x^2)}
\end{align}$$
From here, the rest is pretty easy to show; just perform a partial-fractions decomposition on $\dfrac{x}{(1-3x+x^2)(1-x)^2}$ to write it as $\dfrac{A+Bx}{(1-3x+x^2)}+\dfrac{C}{1-x}+\dfrac{D}{(1-x)^2}$.  The functions $\dfrac{1}{1-x}$ and $\dfrac{1}{(1-x)^2}$ are generating functions for the constant sequence $c_n=1$ and the linear sequence $d_n=(n+1)$, respectively, so you'll be able to translate the coefficients $C$ and $D$ back to those terms in your sequence.
