Fibonacci identity using generating function There are many nice ways of showing that $f_0^2+f_1^2+\cdots+f_n^2=f_{n+1}f_n$. I was wondering if there is a way of showing this using the generating function $F(x)=\frac{1}{1-x-x^2}=\sum_{i\geq0}f_ix^i$. In other words, is there any operation (perhaps the Hadamard product) that can be applied to $F(x)$ that would yield the identity above?
What about other identities that involve sums and squares, like $f_1f_2+\cdots +f_nf_{n+1}=f_{n+1}^2$ for $n$ odd?
 A: Apply the technique described in this blog post to $F(x) F(y)$, then to $x F(x) F(y)$.  The key observation here is that one can compute Hadamard products of rational functions using the residue theorem.
A: Here are the details on proving these identities using Hadamard products of generating functions. (You can find explanations of how to compute Hadamard products of rational functions here.)
I'll write $U(x)*V(x)$ for the Hadmard product of $U(x)$ and $V(x)$:
$$\sum_{n=0}^\infty u_n x^n *\sum_{n=0}^\infty v_n x^n
  =\sum_{n=0}^\infty u_n v_n x^n.$$
Let 
$$F=\sum_{n=0}^\infty f_n x^n =\frac{1}{1-x-x^2}$$
and let 
$$F_1 = \sum_{n=0}^\infty f_{n+1}x^n = (F-1)/x = \frac{1-x}{1-x-x^2}.$$
Then we have
$$
\sum_{n=0}^\infty f_{n}^2x^n=F*F=\frac{1-x}{1-2x-2x^2+x^3}
$$
and
$$
\sum_{n=0}^\infty f_{n}f_{n+1}x^n = F*F_1 = \frac{1}{1-2x-2x^2+x^3}.
$$
Thus $(F*F)/(1-x) =F*F_1$.
This proves the first identity $f_0^2 + \cdots + f_n^2 = f_{n+1}f_n$.
The second identity is stated incorrectly. It should be 
$$f_0f_1 + f_1f_2 +\cdots +f_n f_{n+1}=
\begin{cases}
f_{n+1}^2,&\text{if $n$ is even}\\
f_{n+1}^2 -1,&\text{if $n$ is odd}.
\end{cases}
$$
The generating function for the left side is 
$$\frac{F*F_1}{1-x}=\frac{1}{(1-x)(1-2x-2x^2+x^3)}.$$
We have 
$$\sum_{n=0}^\infty f_{n+1}^2 x^n = F_1*F_1= \frac{1+2x-x^2}{1-2x-2x^2+x^3}$$
and we find that 
$$\frac{F*F_1}{1-x} -F_1*F_1 = -\frac{x}{1-x^2},$$
which proves the corrected second identity.
