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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.

Grade:12

1 Answers

mycroft holmes
272 Points
7 years ago
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}}

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