Evaluate $$\sum_{i=0}^{504} C(2017,4i +1).$$
So what I did is that I computed similar expressions like instead of $2017$ I took $5$, $9$, $13$ ( $4m + 1$ also max value is $\frac {k-1}4 $ ) and I get $$\sum_{i=0}^{\frac {k-1}4} C(k,4i+1)= 2^{k-2} + (-1)^{\frac {k-1}4}\cdot 2^{\frac {k-3}2}.$$According to answer it is true but is there a more elegant solution?
For $r=0,1,2,3$, let $$S_r:=\sum_{k=0}^{N} \binom{4N+1}{4k +r}.$$ We have that $$S_1=\sum_{k=0}^{N} \binom{4N+1}{4(N-k) +1}=\sum_{k=0}^{504} \binom{4N+1}{4N+1-4k}=\sum_{k=0}^{N} \binom{4N+1}{4k}=S_0.$$ Moreover $((1+i)/\sqrt{2})^{4N+1}=(-1)^N(1+i)/\sqrt{2}$ and $(1+1)^{4N+1}=2^{4N+1}$ imply that $$S_0+iS_1-S_2-iS_3=(-1)^N2^{2N}(1+i)\quad\text{and}\quad S_0+S_1+S_2+S_3=2^{4N+1}.$$ Hence $$\begin{cases} S_0=S_1\\ S_0-S_2=(-1)^N2^{2N}\\ S_1-S_3=(-1)^N2^{2N}\\ S_0+S_1+S_2+S_3=2^{4N+1} \end{cases}$$ and by solving the linear system we obtain $$S_0=S_1=2^{4N-1}+(-1)^N2^{2N-1}\quad\text{and}\quad S_2=S_3=2^{4N-1}-(-1)^N2^{2N-1}.$$