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 $-2 + 49 = 47$.
We'll think of the summand as counting (up to sign) even-sized subsets of $\{1, \dots, 99\}$.
Consider the following operation on such sets: Find the smallest $j$ such that $2j-1$ and $2j$ are either both in the subset or both not in the subset. Then either take both out (if they are already in) or put both in (if they are not currently in). If no such $j$ exists, then leave the subsets alone.
This operation is an involution (applying it twice leaves you with the same set you started with), so it'll partition the subsets into pairs and fixed points. For each pair, the subset sizes differ by $2$. This means one subset of the pair contributes $+1$ to the sum, while the other contributes $-1$, and they cancel each other. So all that are left are the fixed points.
Each fixed point contains exactly one of $\{2j-1, 2j\}$ for each $j$, as well as $99$ (to make the subset size even). This means there's $2^{49}$ fixed points, each contributing $-1$ to the final sum.