Question

The variance of the first $n$ natural number is

• A. $\frac{1}{12}\left(n^2-1\right)$
• B. $\frac{1}{6}\left(n^2-1\right)$
• C. $\frac{1}{6}\left(n^2+1\right)$
• D. $\frac{1}{12}\left(n^2+1\right)$

Answer 1

The correct option is A $\frac{1}{12}\left(n^2-1\right)$

Variance $\left({\sigma }^2\right)=\frac{1}{n}{\sum }_{i=1}^n(x_i-\overline{x}{)}^2$

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

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

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

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

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

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

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

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