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$$2^n \cos \left (\frac{n \pi}{2} \right )=\sum_{k=0}^{n} (-1)^k \binom{2n}{2k}$$

I expanded the LHS and got $$\binom{2n}{0}-\binom{2n}{2}+\binom{2n}{4}-\binom{2n}{6}+\cdots+(-1)^{n}\binom{2n}{2n}$$

Is this a telescoping series? Where can I go from here?

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I would break this up into 3 cases: $n$ odd, $n=2 k$ where $k$ is even, and $n=2 k$ where $k$ is odd. Prove each case by induction. –  Ron Gordon May 8 '13 at 9:28
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Hint: What is $(1+i)^{2n}$ and $(1-i)^{2n}$ by binomial theorem? –  achille hui May 8 '13 at 9:28
    
Convergence is usually associated with infinite series, not finite sums. Perhaps a better title would be "Prove that a binomial sum equals a trigonometric expression" –  robjohn May 8 '13 at 9:48

2 Answers 2

up vote 5 down vote accepted

$$ \begin{align} \sum_{k=0}^n\binom{2n}{2k}x^{2k} &=\frac12\left(\sum_{k=0}^{2n}\binom{2n}{k}x^k+\sum_{k=0}^{2n}\binom{2n}{k}(-x)^k\right)\\[6pt] &=\frac{(1+x)^{2n}+(1-x)^{2n}}{2} \end{align} $$ Plug in $x=i$ to get $$ \begin{align} \sum_{k=0}^n(-1)^k\binom{2n}{2k} &=\frac{(1+i)^{2n}+(1-i)^{2n}}{2}\\ &=\frac{(2i)^n+(-2i)^n}{2}\\ &=2^n\frac{e^{in\pi/2}+e^{-in\pi/2}}{2}\\[9pt] &=2^n\cos(n\pi/2) \end{align} $$

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An idea:

$$\cos\frac{n\pi}2=\begin{cases}\;0&,\;\;\text{if}\;\;\;n=1\pmod 2\\{}\\\!\!\!-1&,\;\;\text{if}\;\;\;n=2\pmod 4\\{}\\\;1&,\;\;\text{if}\;\;\;n=0\pmod 4\end{cases}$$

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