The variance of first $n$natural numbers is 

Solution:

We have,

$\begin{array}{l}{\sigma }^2=\frac{1}{n}\sum _{i=1}^n x_i^2-{\left(\frac{1}{n}\sum _{i=1}^n x_i\right)}^2\\ \Rightarrow {\sigma }^2=\frac{1}{n}\left(1^2+2^2+\dots +n^2\right)-{\left(\frac{1}{n}(1+2+\dots +n)\right)}^2\\ \Rightarrow {\sigma }^2=\frac{1}{n}\times \frac{n(n+1)(2n+1)}{6}-{\left(\frac{n+1}{2}\right)}^2=\frac{n^2-1}{12}\end{array}$