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.

Problems in Mathematics of KöMaL, May 2009

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

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