Identity involving binomial coefficients I might have written this in a needlessly cumbersome way, but I want to prove that for odd positive integers $n$, $$\sum_{k\ odd}^{n}\binom{2n+1}{2k}=\begin{cases}
      \binom{2^n+1}{2}, & \text{if}\ n\ \text{mod}\ 4 =1\\
      \binom{2^n}{2}, & \text{if}\ n\ \text{mod}\ 4 =3
    \end{cases}$$
I have tested these identities and they should hold in general. Thank you very much in advance!
Edit:
I found the following formula in the wikipedia article for binomial coefficients, under multisections of sums (https://en.wikipedia.org/wiki/Binomial_coefficient#Multisections_of_sums) which works nicely.$$\binom{n}{2}+\binom{n}{6}+\binom{n}{10}+\cdots=\frac{1}{2}(2^{n-1}-2^{\frac{n}{2}}\cos\frac{n\pi}{4})$$
It would be great if someone can provide a reference or proof for this. Or if there is a much faster way of getting at the same result then please ignore this entirely.

Hint: Use the binomial formula for $(1+1)^{2n+1}$, $(1-1)^{2n+1}$, $(1+i)^{2n+1}$ and $(1-i)^{2n+1}$:
$$A=(1+1)^{2n+1}=\binom{2n+1}{0}+\binom{2n+1}{1}+\binom{2n+1}{2}+\binom{2n+1}{3}+\cdots$$
$$B=(1-1)^{2n+1}=\binom{2n+1}{0}-\binom{2n+1}{1}+\binom{2n+1}{2}-\binom{2n+1}{3}+\cdots$$
$$C=(1+i)^{2n+1}=\binom{2n+1}{0}+\binom{2n+1}{1}i-\binom{2n+1}{2}-\binom{2n+1}{3}i+\cdots$$
$$D=(1-i)^{2n+1}=\binom{2n+1}{0}-\binom{2n+1}{1}i-\binom{2n+1}{2}+\binom{2n+1}{3}i+\cdots$$
so the requested sum is:
$$\binom{2n+1}{2}+\binom{2n+1}{6}+\cdots=\frac{1}{4}(A+B-C-D)$$
All it takes is to calculate $\frac{1}{4}(A+B-C-D)=\frac{1}{4}\left(2^{2n+1}-(1+i)^{2n+1}-(1-i)^{2n+1}\right)$. That can be done for the cases $n=4k+1$ and $n=4k+3$ separately.
For example, for $n=4k+1$, we have:
$$\frac{1}{4}\left(2^{2n+1}-(1+i)^{8k+3}-(1-i)^{8k+3}\right)=\frac{1}{4}\left(2^{2n+1}-2^{4k}\left((1+i)^3-(1-i)^3\right)\right)=\frac{1}{4}\left(2^{2n+1}+2^{4k+2}\right)=\frac{1}{2}(2^{2n}+2^n)=\binom{2^n+1}{2}$$
Above, we used the fact that $(1+i)^8=(1-i)^8=2^4$, and also $(1+i)^3+(1-i)^3=-4$.
The case $n=4k+3$ is similar.
