<s>[INST] 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.
 [/INST] Preliminaries
Note that if $n\mid m$, then
$$
\begin{align}
\frac1n\sum_{k=0}^{n-1}e^{2\pi ikm/n}
&=\frac1n\sum_{k=0}^{n-1}1\\[6pt]
&=1
\end{align}
$$
and that if $n\nmid m$, then
$$
\begin{align}
\frac1n\sum_{k=0}^{n-1}e^{2\pi ikm/n}
&=\frac1n\frac{e^{2\pi im}-1}{e^{2\pi im/n}-1}\\[6pt]
&=0
\end{align}
$$
Thus,
$$
\frac1n\sum_{k=0}^{n-1}e^{2\pi ikm/n}=[\,n\mid m\,]
$$
where $[\dots]$ are Iverson Brackets.
Therefore, setting $n=4$, we get
$$
\begin{align}
[\,k\equiv0\pmod4\,]
&=1^k+i^k+(-1)^k+(-i)^k\\
[\,k\equiv2\pmod4\,]
&=1^{k-2}+i^{k-2}+(-1)^{k-2}+(-i)^{k-2}\\
&=1^k-i^k+(-1)^k-(-i)^k
\end{align}
$$

First Question
$$
\begin{align}
\sum_{k=0}^n\frac{\overbrace{1^k-i^k+(-1)^k-(-i)^k}^{[\,k\equiv2\pmod4\,]}}4\binom{2n+1}{k}
&=\frac{2^{2n+1}-(1+i)^{2n+1}+0^{2n+1}-(1-i)^{2n+1}}4\\
&=2^{2n-1}-2^{n-\frac12}\cos\left(\frac{(2n+1)\pi}4\right)\\[6pt]
&=2^{2n-1}-2^{n-1}\left(\cos\left(\frac{n\pi}2\right)-\sin\left(\frac{n\pi}2\right)\right)\\[6pt]
&=2^{2n-1}+2^{n-1}(-1)^{\left\lfloor\frac{n-1}2\right\rfloor}\\
&=\frac{2^n\left(2^n+(-1)^{\left\lfloor\frac{n-1}2\right\rfloor}\right)}2\\
&=\bbox[5px,border:2px solid #C0A000]{\left\{\begin{array}{}
\binom{2^n+1}{2}&\text{if }n\equiv1,2\pmod4\\
\binom{2^n}{2}&\text{if }n\equiv0,3\pmod4
\end{array}\right.}
\end{align}
$$

Second Question
$$
\begin{align}
\sum_{k=0}^n\frac{\overbrace{1^k-i^k+(-1)^k-(-i)^k}^{[\,k\equiv2\pmod4\,]}}4\binom{n}{k}
&=\frac{2^n-(1+i)^n+0^n-(1-i)^n}4\\
&=\bbox[5px,border:2px solid #C0A000]{2^{n-2}-2^{\frac n2-1}\cos\left(\frac{n\pi}4\right)+\frac{[n=0]}4}
\end{align}
$$
</s>