Question

Write the variance of first n natural numbers.

Answer 1

​Sum of first n natural numbers $=\frac{n\left(n+1\right)}{2}$


Mean, $\overline{X}=\frac{\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{the}\mathrm{observations}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{observations}}$
$=\frac{{\displaystyle \frac{n\left(n+1\right)}{2}}}{n}=\frac{n+1}{2}$



$$\begin{gathered} \therefore {\sigma }^2=\frac{\sum {\left(x_i-\overline{X}\right)}^2}{n}=\frac{\sum {\left(x_i-{\displaystyle \frac{n+1}{2}}\right)}^2}{n} \\ =\frac{1}{n}\sum \left[{x_i}^2-x_i\left(n+1\right)+{\left(\frac{n+1}{2}\right)}^2\right] \\ =\frac{n\left(n+1\right)\left(2n+1\right)}{6n}-\left[\frac{n\left(n+1\right)}{2}\right]\left(\frac{n+1}{n}\right)+\frac{{\left(n+1\right)}^2}{4n}\times n \\ =\frac{\left(n+1\right)\left(2n+1\right)}{6}-\frac{{\left(n+1\right)}^2}{2}+\frac{{\left(n+1\right)}^2}{4} \\ =\frac{\left(n+1\right)\left(n-1\right)}{12}=\frac{\left(n^2-1\right)}{12} \end{gathered}$$