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

I've tried expanding the binomial coefficient in its factorial form and can't seem to get to manipulate it in a way that solves the expression.

$C_{99}^{2k}=\frac{99!}{(99-2k)!(2k)!}=\frac{[2(49)+1]!}{[2(49-k)+1]!(2k)!}$

Is this a step in the right direction? I've written the 99 in form of 49 so it wil match with the binomial upper limit. So it can be finally written in the form of $(x-1)^{49}$. Correct?

Tip: The answer should be $\pm2^{n}$

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Since$$(1+i)^n=\binom{n}{0}+\binom{n}{1}i-\binom{n}{2}-\binom{n}{3}i+\binom{n}{4}+\binom{n}{5}i+\cdots,$$ we can see $$\binom{n}{0}-\binom{n}{2}+\binom{n}{4}-\binom{n}{6}+\cdots$$ is equal to the real part of $(1+n)^n$. Since $$1+i=\sqrt 2(\cos(\pi/4)+i\sin(\pi/4)),$$ we have $$\binom{n}{0}-\binom{n}{2}+\binom{n}{4}-\binom{n}{6}+\cdots=2^{n/2}\cos(n\pi/4).$$ Set $n=99$ for your question.

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

as $-1=i^2\implies(-1)^k=i^{2k}$

Check for $(a+b)^{2n+1}+(a-b)^{2n+1}$

Then set $a=1,b=i$

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  • $\begingroup$ Is it possible without the use of complex numbers? $\endgroup$ – user167289 Oct 3 '14 at 9:50
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    $\begingroup$ @user167281, In my belief no as the $r$th term $$\binom{99}{2r}i^{2r}$$ $\endgroup$ – lab bhattacharjee Oct 3 '14 at 9:53
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Starting from lab bhattacharjee's answer, and in the same spirit as mathlove, you could notice that $$1+i=\sqrt 2 e^{i \pi/4}$$ $$1-i=\sqrt 2 e^{-i \pi/4}$$ and use Moivre theorem. So $$(1+i)^{99}+(1-i)^{99}=2(\sqrt 2)^{99} \cos(\frac{99\pi}{4})=-\frac{2(\sqrt 2)^{99}}{\sqrt 2}=-2^{50}=-1125899906842624$$

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