Let $∑_{n=0}^∞c_n z^n $ be a representation for the function $\frac{1}{1-z-z^2 }$. Find the coefficient $c_n$ Let $∑_{n=0}^∞c_n z^n $ be a power series representation for the function $\frac{1}{1-z-z^2 }$. Find the coefficient $c_n$ and radius of convergence of the series.
Clearly this is a power series with center $z_0=0$, and $f(z)=\frac{1}{1-z-z^2 }$ is analytic, because it's represented by a power series. I also know that
$$c_n =\frac{1}{n!} f^{(n)}(0)$$
but this doesn't get me anywhere, I also try the special case of Taylor series, but nothing look like this. I wonder if any one would give  me a hint please.

Represent it as a geometric series:
$$\frac{1}{1 - z - z^2} = \sum_{k=0}^{\infty} z^k(1 + z)^k$$
The $c_n$ coefficient is effected by the terms $k = \lceil n/2\rceil$ to $k = n$. In particular,
$$c_n = \sum_{k = \lceil n/2 \rceil}^{n} {k  \choose n - k}$$
Can you show that these coefficients are the Fibonacci numbers? (Hint: The fibonacci sequence are the shallow diagonals on Pascal's triangle)
