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Problem A. 482. (May 2009)

A. 482. Let n be a positive integer. Prove that \sum_{k=-n}^n {(-1)}^k
\left(\cos\frac{k\pi}{2n+1}\right)^{2n+1} > 0.

(5 pont)

Deadline expired on June 15, 2009.


Solution. Let \varepsilon=e^{\pi i/(2n+1)}=\cos\frac{\pi}{2n+1}+i\sin\frac{\pi}{2n+1}.


\sum_{k=-n}^n (-1)^k \left(\cos\frac{k\pi}{2n+1}\right)^{2n+1} =
\sum_{k=-n}^n (-1)^k \left(\frac{\varepsilon^k+\varepsilon^{-k}}2\right)^{2n+1} =


= \frac1{2^{2n+1}}\sum_{k=-n}^n (-1)^k \big(\varepsilon^k+\varepsilon^{-k}\big)^{2n+1} =
\frac1{2^{2n+1}}\sum_{k=-n}^n (-1)^k \sum_{\ell=0}^{2n+1}\binom{2n+1}{\ell}
\big(\varepsilon^k\big)^\ell\big(\varepsilon^{-k}\big)^{2n+1-\ell} =


= \frac1{2^{2n+1}}\sum_{\ell=0}^{2n+1}\binom{2n+1}{\ell}
\sum_{k=-n}^n \big(-\varepsilon^{2\ell-(2n+1)}\big)^k
= \frac1{2^{2n+1}}\sum_{\ell=0}^{2n+1}\binom{2n+1}{\ell}
\sum_{k=-n}^n \big(\varepsilon^{2\ell}\big)^k.

The sequence \big(\varepsilon^{2\ell}\big)^k is geometric sequence with quotient \varepsilon^{2\ell}. In the cases \ell=0 and \ell=2n+1, the quitient is 1 and the last sum equals 2n+1. If 0<\ell<2n+1, then


\sum_{k=-n}^n \big(\varepsilon^{2\ell}\big)^k =
\varepsilon^{-n\ell}\frac{(\varepsilon^{2\ell})^{2n+1}-1}{\varepsilon^{2\ell}-1}=0.

Therefore,


\sum_{k=-n}^n (-1)^k \left(\cos\frac{k\pi}{2n+1}\right)^{2n+1} =
\frac1{2^{2n+1}} \left(\binom{2n+1}{0}+\binom{2n+1}{2n+1}\right)(2n+1) =
\frac{2n+1}{2^{2n}}.

We proved that


\sum_{k=-n}^n (-1)^k \left(\cos\frac{k\pi}{2n+1}\right)^{2n+1} =
\frac{2n+1}{2^{2n}},

which is positive.


Statistics:

4 students sent a solution.
5 points:Backhausz Tibor, Nagy 235 János, Nagy 314 Dániel, Tomon István.

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