I'm looking for tips on how to prove the following identity for all $n \in \mathbb{N}$.
$$\sum^{\left \lfloor{\frac{n}{2}}\right \rfloor}_{k=0}\begin{pmatrix}n\\2k\end{pmatrix}=2^{n-1}$$
I know that from the binomial theorem I can show that $$2^n=(1+1)^n=\sum^n_{k=0}\begin{pmatrix}n\\k\end{pmatrix}1^k1^{n-k}=\sum^n_{k=0}\begin{pmatrix}n\\k\end{pmatrix}$$
So how can I make a connection between these two identities? Is it also possible to make a bijection between the sets representing the left and right or count in both directions?
Hint: try to expand $(1-1)^{n}$. The value of this is zero but also equals to odd binomial coefficient substracted by even binomial coefficient
Another hint: imagine choosing a number of people from $n$ people with condition that the number must be even, then You proceed to ask the people one by one if they want to be chosen or not ($2$ possibilities). When You get to the last people, he has no choice but to join if your group is still odd or he has no choice but to not join if your group is already even.