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 $\log_2{(-c)}$.

Question

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 $\log_2{(-c)}$.

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

Answer 1

$$\sum_{99 \geq k \geq 0, \text{even}} i^k {99 \choose k}$$

$$=\sum_{99 \geq k \geq 0} \frac{(-1)^k+1^k}{2}i ^k{99 \choose k}$$

$$=\frac{1}{2} \sum_{k=0}^{99} i^k {99 \choose k}+\frac{1}{2} \sum_{k=0}^{99} (-i)^k {99 \choose k}$$

$$=\frac{1}{2}(1+i)^{99}+\frac{1}{2}(1-i)^{99}$$

$$=\frac{1}{2}(\sqrt{2})^{99}e^{99 \frac{\pi}{4}i }+\frac{1}{2}(\sqrt{2})^{99} e^{-99 \frac{\pi}{4}i}$$

$$=(\sqrt{2})^{99} \frac{e^{99 \frac{\pi}{4}i} +e^{-99\frac{\pi}{4}i}}{2}$$

$$=\left((\sqrt{2})^{99}\right)\left(\cos (\frac{99}{4}\pi) \right)$$

$$=(2^{49})(\sqrt{2})(\frac{-1}{\sqrt{2}})$$

$$=-2^{49}$$

Answer 2

HINT:

$$2\sum_{r=0}^n\binom{2n+1}{2r}i^{2r}=(1+i)^{2n+1}+(1-i)^{2n+1}$$

Now $1\pm i=\sqrt2e^{\pm i\pi/4}$

Answer 3

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