functions nice one

prove that summation k=1 to k=n-1 (n-k)cos(2kpi)/n = -n/2
for all nâ‰¥3 , n belongs to integer.

2 Answers

1357

Manish Shankar ·2009-02-01 22:07:14

take this as

Real part of (n-k)e^2kpi/n

take the sum of the AG series above...

That will give you the answer... this is a standard method...

341

Hari Shankar ·2009-02-02 02:37:32

We need real part of S = Î£_k=1^n-1 (n-k) Ï‰^k, where Ï‰ = cis(2Ï€/n). Note that Ï‰ⁿ = 1

Now conjugate of S = S' = Î£_k=1^n-1 (n-k) Ï‰'^k = Î£_k=1^n-1 (n-k) (Ï‰^k)'

However (Ï‰^k)' = Ï‰^n-k as Ï‰^kÏ‰^n-k = 1

Hence we have S' = Î£_k=1^n-1 (n-k) (Ï‰^k)' = Î£_k=1^n-1 (n-k) Ï‰^n-k = Î£_k=1^n-1 k Ï‰^k

Thus 2 Re(S) = S + S' = Î£_k=1^n-1 (n-k) Ï‰^k + Î£_k=1^n-1 k Ï‰^k

= n Î£_k=1^n-1 Ï‰^k = -n

Hence Re(S) = -n/2

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