Let 1≤r≤901 \leq r \leq 901≤r≤90 be an odd integer. Evaluate ∑k=0rsin2(kπ2r)\displaystyle \sum_{k=0}^r \sin^2\left(\dfrac{k\pi}{2r}\right)k=0∑rsin2(2rkπ). Report the answer when r=45r = 45r=45.
