I want to prove that $\displaystyle \frac{1}{2\pi} \int_0^{2\pi} (\sum_{k=-n+1}^{n+1} (n-|k|)e^{ik\theta})^2d\theta = \sum_{k=-n+1}^{n+1} (n-|k|)^2$.
I have seen some leads using Parseval relations and Fourier coefficients but I haven't done it in class and I'm looking for a solution with just calculus, not with some theorem.
Note that
$$\begin{align} \int_0^{2\pi}\left(\sum_{k=-n+1}^{n+1}(n-|k|)e^{ik\theta}\right)^2\,d\theta&=\sum_{k=-n+1}^{n+1}(n-|k|)\sum_{j=-n+1}^{n+1}(n-|j|)\underbrace{\int_0^{2\pi}e^{ij\theta}e^{ik\theta}\,d\theta}_{=2\pi \delta_{j,-k}}\\\\ &=\sum_{k=-n+1}^{n+1}(n-|k|)\underbrace{\sum_{j=-n+1}^{n+1}(n-|j|)2\pi \delta_{j,-k}}_{=2\pi (n-|-k|)}\\\\ &=2\pi \sum_{k=-n+1}^{n+1}(n-|k|)^2 \end{align}$$
as was to be shown!