TITLE: How to compute $\sum\limits_{k=0}^n (-1)^k{2n-k\choose k}$?
QUESTION [18 upvotes]: I got stuck at the computation of the sum 
$$
\sum\limits_{k=0}^n (-1)^k{2n-k\choose k}.
$$
I think there is no purely combinatorial proof here since the sum can achieve negative values. Could you give me solution, it seems to involve generating functions.

REPLY [2 votes]: If you want only the result, CAS says
$$
\frac{2\sqrt{3}}{3}(\cos\left( \frac{(4n-1)\pi}{6}\right).
$$