Prove that $\sum_{k=1}^{n-1} (n-k) cos\frac{2πk}{n}=\frac{-n}{2}$ where n $\ge$3 is an integer .
My approach, we need to find real part of the equation. Where k is increasing, i am getting confused. Is there any other method of approaching this question.
Consider the function:
$$f(x) = x^n + x^{n-1} + \dots + x^2 = \frac{x^{n+1} - 1}{x-1} -x-1$$
Then we have $2x + 3x^2 + \dots + nx^{n-1} = f'(x) = \frac{(n+1)x^n(x-1) - (x^{n+1} - 1)}{(x-1)^2} - 1$
Now let $\omega$ be the $n-$th root of unity . Then:
$$\sum_{k=1}^{n-1} (k+1)\omega^k = f'(\omega) = \frac{n}{\omega - 1} - 1$$
Now:
$$\sum_{k=1}^{n-1} (n-k)\omega^k = (n+1)\sum_{k=1}^{n-1}\omega^k - \sum_{k=1}^{n-1} (k+1)\omega^k = (n+1)(-1) - \frac{n}{\omega - 1} + 1 = -n - \frac{n}{\omega - 1}$$
Now it's not hard to conclude that $Re\left( \frac{1}{\omega - 1}\right) = \frac{-1 + \cos \frac{2\pi}{n}}{2-2\cos \frac{2\pi}{n}} = - \frac 12$. So:
$$\sum_{k=1}^{n-1} (n-k) cos\frac{2πk}{n} = Re\left(\sum_{k=1}^{n-1} (n-k)\omega^k\right) = Re\left(-n - \frac{n}{\omega - 1}\right) = - \frac n2$$
