Prove that the sum $$\sum_{k=1}^{n-1} (n-k)\cdot\cos\left(\frac{2k\pi}{n}\right) $$ Is an integer for any $n\geq 3$.
I found this in my textbook but am unable to evaluate this sum. Any help would be appreciated.
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This is your 10th month and 85th question here. You HAVE to $\LaTeX$ there is no excuse. – UserX Nov 08 '14 at 11:00
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Such a sum gives an integer only if $n$ is even. Otherwise, it gives half an odd integer. – Jack D'Aurizio Nov 08 '14 at 11:42
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We have (by replacing $k$ with $n-k$): $$S=\sum_{k=1}^{n-1}(n-k)\cos\frac{2\pi k}{n} = \sum_{k=1}^{n-1}k\cos\frac{2\pi k}{n}$$ and by setting $\omega = \exp\frac{2\pi i}{n}$ it follows that: $$ S = \Re \sum_{k=1}^{n-1} k \omega^k. $$ Since: $$\begin{eqnarray*} (1-\omega)\sum_{k=1}^{n-1}k\omega^k&=&\sum_{k=1}^{n-1}k\omega^k-\sum_{k=1}^{n-1}k\omega^{k+1}=\sum_{k=1}^{n-1}k\omega^k-\sum_{k=2}^{n}(k-1)\omega^{k}\\&=&\omega-(n-1)+\sum_{k=2}^{n-1}\omega^k=-n\end{eqnarray*}$$ we have:
$$ S = \Re\frac{n}{\omega-1}=n\cdot\Re\frac{1}{\left(\cos\frac{2\pi}{n}-1\right)+i\sin\frac{2\pi}{n}} = -\frac{n}{2}.$$
Jack D'Aurizio
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Is there a way to solve this problem without complex numbers? Just curious. – user34304 Nov 08 '14 at 12:32
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$$\sum_{k=1}^{n-1}(n-k)\cos\frac{2k\pi }n$$
$$=n\sum_{k=1}^{n-1}\cos\frac{2k\pi }n-\sum_{k=1}^{n-1}k\cos\frac{2k\pi }n$$
$\displaystyle\sum_{k=1}^{n-1}\cos\frac{2k\pi }n=$Real part of $\displaystyle\sum_{k=1}^{n-1}(e^{\frac{2\pi i}n})^k$
Again, $\displaystyle\sum_{k=1}^{n-1}k r^k=r\sum_{k=1}^{n-1}\frac{d(r^k)}{dr}$
See also : $\sum \cos$ when angles are in arithmetic progression
lab bhattacharjee
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