The variance of first n natural numbers is
n2+112
n2-112
(n+1)(2n+1)6
none of these
We have,
σ2=1n∑i=1n xi2−1n∑i=1n xi2⇒ σ2=1n12+22+…+n2−1n(1+2+…+n)2⇒ σ2=1n×n(n+1)(2n+1)6−n+122=n2−112