Alternating binomial sum over even coefficients. Given a positive integer $n$, I'm looking for a nicer closed form for the expression
$$\sum_{\substack{k=0\\2\mid k}}^n(-1)^{\frac k2}2^k\binom{n}{k}.$$
If it helps, it is fine to assume that $n$ is even. This comes from looking for solutions to
$$x^2+y^2=5^n=(1+2i)^n(1-2i)^n,$$
if there's a nicer way to find solutions I'd be happy to know.

Let $k = 2j$, so the sum becomes
$$\sum_j (-1)^j 2^{2j} \binom{n}{2j}$$
Now in
$$\frac{1}{2} [(1+x)^n + (1-x)^n] = \sum_j \binom{n}{2j} x^{2j}$$
Let $x = 2i$, where $i^2 = -1$.
The result is
$$\frac{1}{2} [(1+2i)^n + (1-2i)^n] = \sum_j (-1)^j 2^{2j}\binom{n}{2j}$$
which is the desired sum. 
Some further simplification is possible by noting $1+2i = \sqrt{5} e^{\alpha i}$ and $1-2i = \sqrt{5} e^{-\alpha i}$, where $\alpha = \tan^{-1} 2$.
