If $c$ is equal to $\binom{99}{0}-\binom{99}{2}+\binom{99}{4}-\binom{99}{6}+\cdots+\binom{99}{96}-\binom{99}{98},$ then find $\log_2{(-c)}$. If $c$ is equal to $$\binom{99}{0}-\binom{99}{2}+\binom{99}{4}-\binom{99}{6}+\cdots+\binom{99}{96}-\binom{99}{98},$$then find $\log_2{(-c)}$.

I think the binomial theorem might help but the bottom numbers are skipping by 2's.  How would I apply it now?

$$\sum_{99 \geq k \geq 0, \text{even}} i^k {99 \choose k}$$
$$=\sum_{99 \geq k \geq 0} \frac{(-1)^k+1^k}{2}i  ^k{99 \choose k}$$
$$=\frac{1}{2} \sum_{k=0}^{99} i^k {99 \choose k}+\frac{1}{2} \sum_{k=0}^{99}  (-i)^k {99 \choose k}$$
$$=\frac{1}{2}(1+i)^{99}+\frac{1}{2}(1-i)^{99}$$
$$=\frac{1}{2}(\sqrt{2})^{99}e^{99 \frac{\pi}{4}i }+\frac{1}{2}(\sqrt{2})^{99} e^{-99 \frac{\pi}{4}i}$$
$$=(\sqrt{2})^{99} \frac{e^{99 \frac{\pi}{4}i} +e^{-99\frac{\pi}{4}i}}{2}$$
$$=\left((\sqrt{2})^{99}\right)\left(\cos (\frac{99}{4}\pi) \right)$$
$$=(2^{49})(\sqrt{2})(\frac{-1}{\sqrt{2}})$$
$$=-2^{49}$$
