<s>[INST] Binomial sum identity Everyone knows that $\sum{n \choose 2k}=\sum{n \choose 2k+1}$
My question is follwing
What is the clear form of $\sum_{k=0} (-1)^k{n \choose 2k}$ and $\sum_{k=0} (-1)^k{n \choose 2k+1}$
For example, my exercise is
"calculate a+b where a=$\sum_{k=0}^{25} (-1)^k{50 \choose 2k}$ & b=$\sum_{k=0}^{24} (-1)^k{50 \choose 2k+1}$"
I found that it is relative to number 4. 
If n has a form 4k+2 then a=0 and b=$(-1)^k2^{2k+1}$ or If n has a form 4k then b=0 and a=$(-1)^k2^{2k}$
But it's just my hypothesis.(I just wrote it when n=1,2,3,...,10 and found a rule)
 [/INST] Start with
$$
(1 + i)^{n} 
=
\sum_{k=0}^{n} \binom{n}{k} i^{k}
=
\sum_{h} \binom{n}{2 h} (-1)^{h} 
+
i \sum_{h} \binom{n}{2 h + 1} (-1)^{h},
$$
and then note that
\begin{align}
(1 + i)^{n} 
&= 
\sqrt{2}^{n} \cdot \left( \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right)^{n}
\\&=
\sqrt{2}^{n} \cdot \left( \cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right) \right)^{n}
\\&=
\sqrt{2}^{n} \cdot \left( \cos\left(n \frac{\pi}{4}\right) + i \sin\left(n \frac{\pi}{4}\right) \right).
\end{align}

Note that this extends the argument for the case you mentioned
$$
0 = (1 + (-1))^{n} 
=
\sum_{h} \binom{n}{2 h} 
-
\sum_{h} \binom{n}{2 h + 1}.
$$

The download of generatingfunctionology is to be recommended. 
</s>