I have to prove that:
$$\sum_{k=0}^{(\lfloor \frac{n}{2} \rfloor)} { n \choose 2k} = 2^{n-1}$$
in two ways: A combinatorial way, and with the help of the Pascal Pyramid.
For the combinatorial way I thought about possibilities of choosing a set with an even number of elements, but the $$n \choose 0$$ ruins it all!
Any help will be appreciated