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.
 A: 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.
A: 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
$$
\begin{align}
\sum_{k=0}^n\frac{\overbrace{1^k-i^k+(-1)^k-(-i)^k}^{[\,k\equiv2\pmod4\,]}}4\binom{n}{k}
&=\frac{2^n-(1+i)^n+0^n-(1-i)^n}4\\
&=\bbox[5px,border:2px solid #C0A000]{2^{n-2}-2^{\frac n2-1}\cos\left(\frac{n\pi}4\right)+\frac{[n=0]}4}
\end{align}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{k\ odd}^{n}{2n + 1 \choose 2k} & =
\sum_{k = 0}^{\infty}{2n + 1 \choose 2\bracks{2k + 1}} =
\sum_{k = 1}^{\infty}{2n + 1 \choose 2k}\,{1 - \pars{-1}^{k} \over 2} =
{1 \over 2}\sum_{k = 1}^{\infty}{2n + 1 \choose 2k}\pars{1 - \ic^{2k}}
\\[5mm] & =
{1 \over 2}\sum_{k = 2}^{\infty}{2n + 1 \choose k}
\pars{1 - \ic^{k}}{1 + \pars{-1}^{k} \over 2}
\\[5mm] & =
{1 \over 4}\sum_{k = 2}^{\infty}{2n + 1 \choose k} +
{1 \over 4}\sum_{k = 2}^{\infty}{2n + 1 \choose k}\pars{-1}^{k}
\\[2mm] &
\underbrace{-{1 \over 4}\sum_{k = 2}^{\infty}{2n + 1 \choose k}\ic^{k} -
{1 \over 4}\sum_{k = 2}^{\infty}{2n + 1 \choose k}\pars{-\ic}^{k}}
_{\ds{-\,{1 \over 2}\,\Re\sum_{k = 2}^{\infty}{2n + 1 \choose k}\ic^{k}}}
\\[5mm] & =
{1 \over 4}\bracks{\pars{1 + 1}^{2n + 1} - 1 - \pars{2n + 1}} +
{1 \over 4}\bracks{\pars{1 - 1}^{2n + 1} - 1 + \pars{2n + 1}}
\\[2mm] &
-{1 \over 2}\,\Re\bracks{\pars{1 - \ic}^{2n + 1} - 1 - \pars{2n + 1}\ic}
\\[5mm] & =
2^{2n - 1} - {1 \over 2} -
{1 \over 2}\,\Re\bracks{\root{2}\expo{-\pi\ic/4}}^{2n + 1} + {1 \over 2} =
2^{2n - 1} - {1 \over 2}\,\Re\bracks{2^{n + 1/2}\expo{-\pars{2n + 1}\pi\ic/4}}
\\[5mm] & =
2^{2n - 1} - 2^{n - 1/2}\cos\pars{\bracks{2n + 1}\pi \over 4}
\\[5mm] & =
2^{2n - 1} - 2^{n - 1/2}\bracks{\cos\pars{n\pi \over 2}\cos\pars{\pi \over 4} -
\sin\pars{n\pi \over 2}\sin\pars{\pi \over 4}}
\\[5mm] & =
2^{2n - 1}\bracks{1 - \cos\pars{n\pi \over 2} + \sin\pars{n\pi \over 2}}
\\[5mm] & =
\bbx{\left\{\begin{array}{lcl}
\ds{2^{2n - 1}\bracks{1 - \pars{-1}^{n/2}}} & \mbox{if} &
\ds{n}\ \mbox{is}\ even
\\[2mm]
\ds{2^{2n - 1}\bracks{1 + \pars{-1}^{\pars{n - 1}/2}}} & \mbox{if} &
\ds{n}\ \mbox{is}\ odd
\end{array}\right.}
\end{align}
