I am trying to prove the follwowing: $$\sum_{k=0}^n F_kF_{n-k}=\sum_{k=0}^n(k+1)F_{k+1}(-2)^{n-k}$$
where $F_k = F_{k-1} + F_{k-2}$ with $F_0 = F_1=1$ are the Fibonacci numbers.
Now the left hand side of the identity is simply the cauchy product of the $n$-th coefficient of the generating function multiplied by itself, meaning the $n$-th coefficient of $$F(z)\cdot F(z)$$ where $F(z)=\sum_{k\ge 0} F_k z^k$. I am not sure however, if this insight really helps a lot. The only thing I know about the coefficients are its explicit form $$F_k=\sum_{k\ge 0}\binom{n-k}{k}$$ but this did not lead to any results as I would be left with $$\sum_{k=0}^n F_kF_{n-k}=\sum_{k=0}^n\sum_{l\ge 0 }\binom{n-l}{l}\sum_{j\ge 0}\binom{n-j}{j}$$
I would appreciate any hints or ideas.
Well, if you let $$A_n=\sum_{k=0}^n(k+1)F_{k+1}(-2)^{n-k}$$ then \begin{align} \sum_{n=0}^\infty A_nz^n&=\sum_{k=0}^\infty(k+1)F_{k+1}z^k\times\sum_{j=0}^\infty(-2z)^j\\ &=\frac{d}{dz}\left(F(z)\right)\times\frac1{1+2z} =\frac1{1+2z}\frac{d}{dz}\left(\frac 1{1-z-z^2}\right)\\ &=\frac1{1+2z}\frac{1+2z}{(1-z-z^2)^2}=F(z)^2. \end{align} Another triumph for generating functions!