TITLE: Calculating $\sum_{0\le k\le n/2} \binom{n-k}{k}$
QUESTION [10 upvotes]: I would like to evaluate: 
$$\sum_{0\le k\le n/2}\binom{n-k}{k}$$
Any idea?

REPLY [3 votes]: Consider the generating function
$$f(z) = \sum_{n\ge 0} z^n \sum_{k=0}^{\lfloor n/2 \rfloor} {n-k\choose k}$$
and re-write it in terms of $k$ as follows
$$f(z) = \sum_{k\ge 0} \sum_{\lfloor n/2 \rfloor \ge k} z^n {n-k\choose k}
= \sum_{k\ge 0} \sum_{n\ge 2k} {n-k\choose k} z^n
= \sum_{k\ge 0} \sum_{n\ge 0} {n+k\choose n} z^{n+2k}.$$
Now the Newton binomial can be applied to the inner sum to get
$$f(z) = \sum_{k\ge 0} z^{2k} \frac{1}{(1-z)^{k+1}}
= \frac{1}{1-z}\frac{1}{1-z^2/(1-z)} = \frac{1}{1-z-z^2}.$$
This is immediately seen to be the OGF of $F_{n+1}$ ($n+1$-th Fibonacci number) and we are done.
The same trick was used here at this MSE link.