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How do I find the following sum? $\sum_{k=0}^n(-1)^k{2n\choose2k}$

Tried to simplify it somehow but got nothing less complicated.

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  • $\begingroup$ Maybe the binomial theorem, $(x+y)^n=\sum_{k=0}^n{n\choose k}x^{n-k}y^k$, could help. Maybe with $2n$ instead of $n$, and $x=1$ and $y=-1$ ? Then see if you can manipulate it into what you actually want. $\endgroup$ – Pixel Oct 27 '16 at 14:56
  • $\begingroup$ Look closely at $(1+i)^{2n}$ and $(1-i)^{2n}$. $\endgroup$ – Daniel Fischer Oct 27 '16 at 14:56
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Hint:

\begin{align} \sum\limits_{k=0}^{n} {{2n}\choose{2k}} (-1)^k &= \text{ even terms of } \sum\limits_{k=0}^{2n} {{2n}\choose{k}} i^k \\ &= \text{ real part of } \sum\limits_{k=0}^{2n} {{2n}\choose{k}} i^k \\ &= \text{ real part of } (1+i)^{2n} \end{align}

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Hint. One may observe that, by the binomial theorem, $$ \begin{align} \sum_{k=0}^n(-1)^k{2n\choose2k}&=\sum_{k=0}^n {2n\choose2k}i^{2k} \\\\&=\sum_{k=0}^{2n} {2n\choose2k}i^{2k} \\\\&=\sum_{p=0}^{2n} {2n\choose p}\frac{i^{p}+(-i)^p}2 \\\\&=\dfrac{(1+i)^{2n}+(1-i)^{2n}}{2} \\\\&=(2i)^{n}\frac{1+(-1)^n}2. \end{align} $$

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  • $\begingroup$ Why is your last line not $\dfrac{(1+i)^{2n}+(1-i)^{2n}}{2}$? $\endgroup$ – Henry Oct 27 '16 at 15:03
  • $\begingroup$ @Henry Because $(1+i)^{2n}=\big(\sqrt{2}e^{i\pi/4}\big)^{2n}=(2i)^n$. $\endgroup$ – Olivier Oloa Oct 27 '16 at 15:27
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    $\begingroup$ OK. So you are saying the sum is $2^n$ when $n$ is a multiple of $4$, it is $-2^n$ when $n $ is even but not a multiple of $4$, and is $0$ when $n$ is odd. That works for me. $\endgroup$ – Henry Oct 27 '16 at 15:39

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