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!
 A: 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.
A: It's not a complete answer but I think is useful for proofing this question. In fact, I got the following method from this post.
  We can rewrite the question as follows
$$S=n^2\,F_1^2+ {(n-1)}^2\,F_2^2+\cdots+2^2\,F_{n-1}^2+F_n^2$$
where $F_n$ is the $n$th term of Fibonacci number and $n$ is a natural number. Suppose that 
$$
\left\{
\begin{array}{ccc}
U=n\,F_1^2+ {(n-1)}\,F_2^2+\cdots+2\,F_{n-1}^2+F_n^2&&\, ,\\
\\
V_i=i\, F_{n-i+1}^2+(i-1)\, F_{n-i+2}^2+\cdots+ 2\, F_{n-1}^2+F_n^2&&\, .
\end{array}
\right.
$$
It is easy to see that 
$$
S=\sum_{i=0}^{n-1}\, (U-V_i)
$$
Notice that $V_0=0$. So, the values of $S$ is as follows
$$
S=n\,U-\sum_{i=0}^{n-1}\, V_i
$$
In continue, we obtain the values of the $U$ with the same strategy as shown
$$
\left\{
\begin{array}{ccc}
W=F_1^2+ F_2^2+\cdots+F_{n-1}^2+F_n^2&&\, ,\\
\\
Z_i= F_{n-i+1}^2+ F_{n-i+2}^2+\cdots+  F_{n-1}^2+F_n^2&&\, .
\end{array}
\right.
$$
That we can see
$$
U=\sum_{i=0}^{n-1}\, (W-Z_i)
$$
That results that
$$
U=n\,W-\sum_{i=0}^{n-1}\, Z_i=n\,W-\sum_{i=0}^{n-1}\, (W-(F_1^2+ F_2^2+\cdots+F_{n-i}^2))
$$
$$
U=\sum_{i=0}^{n-1}\, (F_1^2+ F_2^2+\cdots+F_{n-i}^2)=\sum_{i=0}^{n-1}\,F_{n-i}\,F_{n-i+1}=
\sum_{i=2}^{n+1}\,F_{i-1}\,F_{i}
$$
The last relation is based on this fact that $\sum_{i=1}^{n}\,F_i^2=F_n\,F_{n+1}$. By using relations $40$ and $42$ 
of this book, we have 
\begin{eqnarray}
\sum_{i=1}^{2n}\, F_i\,F_{i-1}=F_{2n}^2 &\quad , \quad& \sum_{i=1}^{2n+1}\, F_i\,F_{i-1}=F_{2n+1}^2-1
\end{eqnarray}
For calculating $V_i$, we have the same method as follows 
$$
\left\{
\begin{array}{ccc}
A=F_{n-i+1}^2+ F_{n-i+2}^2+\cdots+F_{n-1}^2+F_n^2&&\, ,\\
\\
B_j= F_{n-j+1}^2+ F_{n-j+2}^2+\cdots+  F_{n-1}^2+F_n^2&&\, .
\end{array}
\right.
$$
that we can see the following relation
$$
V_i=\sum_{j=0}^{i-1}\, (A-B_j)
$$
But obtaining $V_i$ is a bit complicated. Maybe it is clear why almost proofs in the Fibonacci coding and other recursive relations is 
based on induction. Proofing with the direct method is so difficult. I hope this incomplete answer be useful for you.
