Find the value of ∑k=1nsin2(kπ2n+1)\sum_{k=1}^{n} \sin^2(\frac{k\pi}{2n+1})∑k=1nsin2(2n+1kπ).
2n+14\frac{2n+1}{4}42n+1
2n4\frac{2n}{4}42n
n2\frac{n}{2}2n
n+12\frac{n+1}{2}2n+1