Taylor Expansion of complex function $ \frac{1}{1 - z - z^2} $ Knowing that 

$$c_{n+2}=c_{n+1}+c_n, \quad n \geq 0,$$ 

the problem is to prove that we have

$$ \frac{1}{1 - z - z^2}=\sum^{+\infty}_{n=0} c_nz^n.$$


I have tried to transform the function into the form:
$$ A(\frac{1}{z-b_1}+\frac{1}{b_2-z})$$
Then expand $\frac{1}{z-b_1} \text{and} \frac{1}{b_2-z}$
However, the result seems having nothing to do with $c_{n+2}=c_{n+1}+c_n$
Am I wrong? How should I prove this in a elegant way?
 A: Hint:
$$
(1-z-z^2)\sum_{n\ge 0}c_nz^n = 1
$$
What is the coefficient of $z^n$ on the each side?
A: You may start from
$$c_{n+2}=c_{n+1}+c_n, \quad n\geq0. \tag1$$
Multiply $(1)$ by $z^{n+2}$ and sum over $n\geq 0$ then you obtain
$$
\begin{align}
\sum^{+\infty}_{n=0} c_{n+2}z^{n+2}&=\sum^{+\infty}_{n=0} c_{n+1}z^{n+2}+\sum^{+\infty}_{n=0} c_{n}z^{n+2}\\
\sum^{+\infty}_{n=0} c_{n+2}z^{n+2}&=z\sum^{+\infty}_{n=0} c_{n+1}z^{n+1}+z^2\sum^{+\infty}_{n=0} c_{n}z^{n}\\
\sum^{+\infty}_{n=0} c_{n+2}z^{n+2}&=z\sum^{+\infty}_{n=1} c_{n}z^{n}+z^2\sum^{+\infty}_{n=0} c_{n}z^{n}\\
\sum^{+\infty}_{n=0} c_{n}z^{n}-c_0-c_1z&=z\sum^{+\infty}_{n=0} c_{n}z^{n}-c_{0}z+z^2\sum^{+\infty}_{n=0} c_{n}z^{n}\\
\end{align}
$$ and you readily obtain
$$ \frac{1}{1 - z - z^2}=\sum^{+\infty}_{n=0} c_nz^n$$ taking into account that $c_0=1$ and $c_1=1$.
A: use this
$$\frac{8 x^2}{\sqrt{5} \left(\sqrt{5}+1\right)^2 \left(2 x+\sqrt{5}+1\right)}+\frac{8 x^2}{\sqrt{5} \left(\sqrt{5}-1\right)^2 \left(-2 x+\sqrt{5}-1\right)}+x+1$$
