Find the mean and variance for the first $n$ natural numbers

The mean of first $n$ natural numbers is calculated as follows.

Mean $=\frac{\text { Sum of all observations }}{\text { Number of observations }}$

$\therefore$ Mean $=\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}$

Varianvce   $\left( {{\sigma ^2}} \right) = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} $

$ = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left[ {{x_i} - \left( {\frac{{n + 1}}{2}} \right)} \right]}^2}} $

$ = \frac{1}{n}\sum\limits_{i = 1}^n {x_i^2 - \frac{1}{n}\sum\limits_{i = 1}^n {2\left( {\frac{{n + 1}}{n}} \right)} } {x_i} + \frac{1}{n}{\sum\limits_{i = 1}^n {\left( {\frac{{n + 1}}{2}} \right)} ^2}$

$=\frac{1}{n} \frac{n(n+1)(2 n+1)}{6}-\left(\frac{n+1}{n}\right)\left[\frac{n(n+1)}{2}\right]+\frac{(n+1)^{2}}{4 n} \times n$

$=\frac{(n+1)(2 n+1)}{6}-\frac{(n+1)^{2}}{2}+\frac{(n+1)^{2}}{4}$

$=\frac{(n+1)(2 n+1)}{6}-\frac{(n+1)^{2}}{4}$

$=(n+1)\left[\frac{4 n+2-3 n-3}{12}\right]$

$=\frac{(n+1)(n-1)}{12}$

$=\frac{n^{2}-1}{12}$