TITLE: finding the combinatorial sum
QUESTION [5 upvotes]: How to find the sum of combinatorial summation of the following series, where
$C(n, k)$ denotes the number of combinations of $n$ given $k$ are the same?
$$\sum_{k=0}^{n/2} C(n-k, k)$$
Need help on this.

REPLY [6 votes]: Another approach is to look at the generating function
$$
\begin{align}
\sum_n\sum_k\binom{n-k}{k}x^n
&=\sum_n\sum_k\binom{n}{k}x^{n+k}\\
&=\sum_nx^n(1+x)^n\\
&=\frac1{1-x-x^2}\\
&=\frac1{\sqrt5}\left(\frac{\phi}{1-\phi x}+\frac{1/\phi}{1+x/\phi}\right)\\
&=\sum_n\frac{\phi^{n+1}-(-1/\phi)^{n+1}}{\sqrt5}x^n
\end{align}
$$
Thus,
$$
\begin{align}
\sum_k\binom{n-k}{k}
&=\frac{\phi^{n+1}-(-1/\phi)^{n+1}}{\sqrt5}\\
&=F_{n+1}
\end{align}
$$
where $F_n$ is the $n^{\text{th}}$ Fibonacci number.