Determine the value of ∑k=1n−1sin2(kπn)\sum_{k=1}^{n-1} \sin^2\left(\frac{k\pi}{n}\right)∑k=1n−1sin2(nkπ) for an integer n≥2n \geq 2n≥2.
n2\frac{n}{2}2n
n−12\frac{n-1}{2}2n−1
n+12\frac{n+1}{2}2n+1
nnn