Calculate in terms of $n$ the following sum:

$$
\sum_{k \equiv 1[4]}\binom{n}{k}
$$
Fix $n>0$, we define the 4 sums as follows:

$$
S_{i}=\sum_{k \equiv i[4]}\binom{n}{k} \quad i \in\{0,1,2,3\}
$$

Now we will use the formula $(1+x)^{n}=\sum\binom{n}{k} x^{k}$ with $x=1,-1$ and $i$.

$$
\begin{aligned}
S_{0}+S_{1}+S_{2}+S_{3} & =2^{n} \\
S_{0}-S_{1}+S_{2}-S_{3} & =0 \\
S_{0}-S_{2} & =\operatorname{Re}\left((1+i)^{n}\right) \\
S_{1}-S_{3} & =\operatorname{Im}\left((1+i)^{n}\right)
\end{aligned}
$$

Then it suffices to use that $(1+i)^{2}=2 i$ and we have the following cases:

$$
\begin{aligned}
& n=4 k \quad: \quad S_{1}=2^{4 k-2} \\
& n=4 k+1 \quad: \quad S_{1}=2^{4 k-1}+(-1)^{k} 2^{2 k-1} \\
& n=4 k+2 \quad: \quad S_{1}=2^{4 k} \quad+(-1)^{k} 2^{2 k} \\
& n=4 k+3 \quad: \quad S_{1}=2^{4 k+1} \quad+(-1)^{k} 2^{2 k}
\end{aligned}
$$