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

Question

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.

Answer 1

Hint:

It is the real part of $$(1+i)^{99}=\Bigl(\sqrt 2\,\mathrm{e}^{\tfrac{i\pi}4}\Bigr)^{99}.$$