if $F_{n+1}$ is the the $n+1$th Fibonacci number, is it possible to derive the identity
$F_{n+1} = \sum_{k=0}^{n}\binom{n-k}{k}$
by using the generating function $F(x) = \frac{x}{(1-x-x^2)}$?
The only way that I could think of doing this was maybe by showing that
$\frac{F^{(n+1)}(0)}{(n+1)!} = \sum_{k=0}^{n}\binom{n-k}{k}$
and I was able to derive the following expression for the $n$th derivative of $F(x)$,
$\frac{F^{(n)}(x)}{n!} = \frac{x^{n+1} + \sum_{k=0}^{n}F_k\binom{n+1}{n-k}x^{n-k}}{(1-x-x^2)^{n+1}}$
Which will give us $\frac{F^{(n+1)}(0)}{(n+1)!} = F_{n+1}$, but not the binomial sum expression. Is there any way to demonstrating that the coefficient of $x^0$ in the above expression is equal to $\sum_{k=0}^{n}\binom{n-k}{k}$? Or any other way of deriving the same from the generating function?