TITLE: Summation of Fibonacci numbers.
QUESTION [5 upvotes]: Let $f_n$ be the sequence of Fibonacci numbers.
We need to show that
$$\sum_{n\ge0} f_n x^n = \dfrac{1}{1-x-x^2}$$
I remember a solution when we are using the generating functions like:
$f(x) = F_0 + F_1x + F_2 x^2 + F_3x^3$
Then multiply by $x$ and $x^2$ both sides and get $(1-x-x^2)f(x) = F_0 + (F_1-F_0)x=x$ and the result follows.
However I don't quite understand how this works. Also I would like to come up to similar expressions for $\sum_{n\ge0} f_{2n} x^n$  and  $\sum_{n\ge0} f_{2n+1} x^n$.

REPLY [3 votes]: \begin{align}
g(x) & = \dfrac1{1-x-x^2} = \sum_{k=0}^{\infty} (x+x^2)^k = \sum_{k=0}^{\infty} x^k(1+x)^k = \sum_{k=0}^{\infty}x^k \sum_{l=0}^k \dbinom{k}l x^l\\
& = \sum_{k=0}^{\infty} \sum_{l=0}^k \dbinom{k}l x^{k+l} = \sum_{m=0}^{\infty}x^m \left(\sum_{l=0}^m \dbinom{m-l}l\right) = \sum_{m=0}^{\infty} g_{m+1} x^m
\end{align}
Now use the fact that
$$\dbinom{n}r + \dbinom{n}{r+1} = \dbinom{n+1}{r+1}$$ to conclude that
$$g_{m+1} = g_m + g_{m-1}$$
with $g_1 = 1$, $g_0 = 0$.