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.
 A: 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 $\alpha = \tan^{-1} 2$.
A: $$
\begin{align}
\sum_{\substack{k=0\\2\mid k}}^n(-1)^{\frac k2}2^k\binom{n}{k}
&=\operatorname{Re}\left(\sum_{k=0}^n(2i)^k\binom{n}{k}\right)\\
&=\operatorname{Re}\left((1+2i)^n\right)\\[12pt]
&=5^{n/2}\cos\left(n\tan^{-1}(2)\right)
\end{align}
$$
A: If we let $(x+iy) = (1+2i)^k(1-2i)^j$ and $(x-iy) = (1+2i)^{n-k}(1-2i)^{n-j}$,
then we find that the difference, $2iy = 5^j(1+2i)^{k-j} - 5^{n-k}(1-2i)^{k-j}$, will only work for an integer value of $y$ if $j = n-k$ as otherwise there would be real parts on the RHS. 
In this case we have $2iy = 5^j((1+2i)^{k-j} - (1-2i)^{k-j})$
Getting this into polar form will yield $2iy = 5^jr^{k-j}(e^{i\theta(k-j)}-e^{-i\theta(k-j)})$ or $y = 5^jr^{k-j}sin(\theta(k-j))$ for integer values of $k,j$ where $k+j=n$.
Similarly, $x = 5^jr^{k-j}cos(\theta(k-j))$ for respective values of $k,j$
where $r = \sqrt{5}$ and $\theta = \arctan{2}$.
So finding a closed form for $y$ or $x$ is equivalent to finding a closed form for the sine or cosine function from this point on-wards.
