<s>[INST] 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?
 [/INST] 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.:
$$
\begin{gathered}
  \log _2 \left( { - c} \right) = \log _2 \left( { - 2^{\,99/2} \cos \left( {\frac{{99\,\pi }}
{4}} \right)} \right) = \log _2 \left( { - 2^{\,99/2} \cos \left( {\frac{{3\,\pi }}
{4}} \right)} \right) =  \hfill \\
   = \log _2 \left( {2^{\,99/2} \frac{{\sqrt 2 }}
{2}} \right) = \log _2 \left( {2^{\,98/2} } \right) = 49 \hfill \\ 
\end{gathered} 
$$
</s>