$2^n$ binomial theorem How can I prove that 
$$2^n=2\left({n \choose 0}+{n \choose 2}+{n \choose 4}+\dots\right)$$
using the binomial theorem. I've tried expanding $(x-y)^n$ with multiple different values of $x$ and $y$ but nothing pans out. I can see why the above is true when looking at Pascal's triangle as the sum of the $n-1$ row gives $2^{n-1}$ and is indeed the sum above, but I can't seem to pin it down algebraically. Thank you.

Hint: 
$$
(x+y)^n = \sum_{k=0}^n \binom nk x^ky^{n-k}
$$Now change $y\to -y$ and make the sum and difference of both relations.
