Prove that $\sum_{k = 0}^{49}(-1)^k\binom{99}{2k} = -2^{49}$ I am preparing a class on the binomial of Newton. One of the exercises at the end of the chapter turns out to be very hard for me:

Prove that $$\sum_{k = 0}^{49}(-1)^k\binom{99}{2k} = -2^{49}$$

I tried to use a clever way of rewriting the binomial coefficients and also tried to use the binomial of Newton to rewrite $2^{49}$ as $\sum_{k = 0}^{49}\binom{49}{k}$, but with no result.
I also tried to use a proof by induction, but got stuck also.
Any hint would be appreciated.
 A: Hint:
It is the real part of 
$$(1+i)^{99}=\Bigl(\sqrt 2\,\mathrm{e}^{\tfrac{i\pi}4}\Bigr)^{99}.$$
A: Hint:
$$(a+b)^{2k+1}+(a-b)^{2k+1}=?$$
Set $a=1,b=i,b^2=-1$
A: The sum $$\sum_{k=0}^{49}(-1)^k\binom{99}{2k}\tag1$$ counts even subsets of the set $N=\{1,2,\dots,99\}$, except that subsets whose sizes are a multiple of four are counted positively, and those whose size is not a multiple of four (but still even) are counted negatively.
Given an even size subset $S$ of $N$, consider the size of the intersection of $S$ with each of the sets
$$
\{1,2\},\{3,4\},\dots,\{97,98\}
$$
We define the following  involution $f$ on even subsets of $N$. Given $S$, find the smallest $k$ for which $S$ contains either both or neither of $\{2k+1,2k+2\}$. If $S$ contains neither, then $f(S)$ is obtained by adding $2k+1$ and $2k+2$ is $S$. If $S$ contains both, then $f(S)$ is obtained by removing these two elements. Note $f(S)$ still has an even number of elements.
Note that almost all of the even subsets of $N$ are divided into pairs $\{S,f(S)\}$. In each pair, one set has size which is a multiple of four, and the other does not. Therefore, the pair $\{S,f(S)\}$ cancels itself out in the summation in $(1)$, so can be ignored. 
However, $f$ is not actually defined for all elements of $S$. If $S$ contains exactly one of $\{2k+1,2k+2\}$ for all $k=0,1,2,\dots,48$, then you cannot compute $f(S)$. The number of such expectational sets is $2^{49}$ (for each $k$, choose if $S$ has $2k+1$ or $2k+2$), and these are all counted negatively in $(1)$, as their size is $50$. As argued before, this is the only thing which contributes to the sum, so we are done.
