<s>[INST] Sum of $\sum_{k=0}^{49} (-1)^{k} \binom{99}{2k}$ I have a finite sum to evaluate. I do compute the sum using Euler's formula. I would appreciate a combinatorial argument, or an argument using a binomial identity.
Problem
Among all the sums obtained from adding the coordinates to the integral solutions of
\begin{equation*}
\sum_{k=0}^{49} (-1)^{k} \binom{99}{2k} = x^{y} ,
\end{equation*}
which is the biggest?
Solution
According to Euler's formula,
\begin{align*}
&2^{99/2} \left(\frac{-1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}\right) \\
&\qquad = (2^{99/2}) e^{i \frac{3\pi}{4}} \\
&\qquad = 2^{99/2} e^{i \frac{99}{4} \pi} \\
&\qquad = (1 + i)^{99} \\
&\qquad = \sum_{k=0}^{49} (-1)^{k} \binom{99}{2k} + i \sum_{k=0}^{49} (-1)^{k} \binom{99}{2k+1} .
\end{align*}
So,
\begin{equation*}
-2^{49} = -2^{99/2} \left(\frac{1}{\sqrt{2}}\right) = \sum_{k=0}^{49} (-1)^{k} \binom{99}{2k} .
\end{equation*}
Since $-2$ is a prime number, the only integral solutions to the given equation are $(-2, \, 49)$, $(-2^{7}, \, 7)$, and $(-2^{49}, \, 1)$. The biggest sum obtained from adding the coordinates to the integral solutions is $-2 + 49 = 47$.
 [/INST] The way is right.
Really, 
$$\sum\limits_{k=0}^{49}(-1)^k\binom{99}{2k} = \Re (1+i)^{99} = \Re (2i)^{49}(1+i)= -2^{49}.$$
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