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?
 A: $$\sum_{99 \geq k \geq 0, \text{even}} i^k {99 \choose k}$$
$$=\sum_{99 \geq k \geq 0} \frac{(-1)^k+1^k}{2}i  ^k{99 \choose k}$$
$$=\frac{1}{2} \sum_{k=0}^{99} i^k {99 \choose k}+\frac{1}{2} \sum_{k=0}^{99}  (-i)^k {99 \choose k}$$
$$=\frac{1}{2}(1+i)^{99}+\frac{1}{2}(1-i)^{99}$$
$$=\frac{1}{2}(\sqrt{2})^{99}e^{99 \frac{\pi}{4}i }+\frac{1}{2}(\sqrt{2})^{99} e^{-99 \frac{\pi}{4}i}$$
$$=(\sqrt{2})^{99} \frac{e^{99 \frac{\pi}{4}i} +e^{-99\frac{\pi}{4}i}}{2}$$
$$=\left((\sqrt{2})^{99}\right)\left(\cos (\frac{99}{4}\pi) \right)$$
$$=(2^{49})(\sqrt{2})(\frac{-1}{\sqrt{2}})$$
$$=-2^{49}$$
A: HINT:
$$2\sum_{r=0}^n\binom{2n+1}{2r}i^{2r}=(1+i)^{2n+1}+(1-i)^{2n+1}$$
Now $1\pm i=\sqrt2e^{\pm i\pi/4}$
A: Since
$$
\begin{gathered}
  \left( {1 + i\,} \right)^{\,n}  = \sqrt 2 ^{\,n} \left( {\cos \left( {\frac{{n\,\pi }}
{4}} \right) + i\sin \left( {\frac{{n\,\pi }}
{4}} \right)} \right) = \sum\limits_{0\, \leqslant \,k\left( { \leqslant \;n} \right)} {\left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)i^{\,k} }  =  \hfill \\
   = \sum\limits_{0\, \leqslant \,j\left( { \leqslant \;n/2} \right)} {\left( \begin{gathered}
  n \\ 
  2j \\ 
\end{gathered}  \right)i^{\,2j} }  + \sum\limits_{0\, \leqslant \,j\left( { \leqslant \;n/2} \right)} {\left( \begin{gathered}
  n \\ 
  2j + 1 \\ 
\end{gathered}  \right)i^{\,2j + 1} }  =  \hfill \\
   = \sum\limits_{0\, \leqslant \,j\left( { \leqslant \;n/2} \right)} {\left( { - 1} \right)^{\,j} \left( \begin{gathered}
  n \\ 
  2j \\ 
\end{gathered}  \right)}  + i\sum\limits_{0\, \leqslant \,j\left( { \leqslant \;n/2} \right)} {\left( { - 1} \right)^{\,j} \left( \begin{gathered}
  n \\ 
  2j + 1 \\ 
\end{gathered}  \right)}  \hfill \\ 
\end{gathered} 
$$
Then
$$
\sum\limits_{0\, \leqslant \,j\left( { \leqslant \;n/2} \right)} {\left( { - 1} \right)^{\,j} \left( \begin{gathered}
  n \\ 
  2j \\ 
\end{gathered}  \right)}  = \sqrt 2 ^{\,n} \cos \left( {\frac{{n\,\pi }}
{4}} \right)
$$
i.e.:
$$
\begin{gathered}
  \log _2 \left( { - c} \right) = \log _2 \left( { - 2^{\,99/2} \cos \left( {\frac{{99\,\pi }}
{4}} \right)} \right) = \log _2 \left( { - 2^{\,99/2} \cos \left( {\frac{{3\,\pi }}
{4}} \right)} \right) =  \hfill \\
   = \log _2 \left( {2^{\,99/2} \frac{{\sqrt 2 }}
{2}} \right) = \log _2 \left( {2^{\,98/2} } \right) = 49 \hfill \\ 
\end{gathered} 
$$
