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Prove that summation of (n-k)cos(2kπ/n) from k=1 to n-1 is equal to -n/2, where n≥3 is an integer.
Prove that summation of (n-k)cos(2kπ/n) from k=1 to n-1 is equal to -n/2, where n≥3 is an integer.


4 years ago

mycroft holmes
272 Points
							Note that $\cos \left(\frac{2(n-k) \pi}{n} \right ) = \cos \left(\frac{2k \pi}{n} \right )$ Hence we have $2S = \sum_{k=1}^{n-1} (n-k)\cos \left(\frac{2k \pi}{n} \right ) + \sum_{k=1}^{n-1} k \cos \left(\frac{2(n-k) \pi}{n} \right )$ $= \sum_{k=1}^{n-1} (n-k)\cos \left(\frac{2k \pi}{n} \right ) + \sum_{k=1}^{n-1} k \cos \left(\frac{2k \pi}{n} \right )$ $= \sum_{k=1}^{n-1} n\cos \left(\frac{2k \pi}{n} \right ) = n\sum_{k=1}^{n-1} \cos \left(\frac{2k \pi}{n} \right )$ $\zeta= e^{\frac{2 \pi i}{n} }$. Then we know that since it is an nth root of unity, we must have  $1+\zeta +\zeta^2+...+\zeta^{n-1} = 0$ or $\zeta +\zeta^2+...+\zeta^{n-1} =-1$ Comparing real parts we see that $\sum_{k=1}^{n-1} \cos \left(\frac{2k \pi}{n} \right )=-1$ Thus we get that $2S = n \sum_{k=1}^{n-1} \cos \left(\frac{2k \pi}{n} \right )=-n$ and hence we have $\boxed{S=-\frac{n}{2}}$

4 years ago
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