Question

The variance of first n natural numbers is :

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

Answer 1

Answer: (D)

We know that variance ${\sigma }^2=\frac{\sum x_1^2}{n}-{\left(\frac{\sum x_1}{n}\right)}^2$
First $n$ natural numbers are:

$\therefore \sum x_i^2=\frac{n(n+1)(2n+1)}{6}$
and $\sum x_i=\frac{n(n+1)}{2}$
${\sigma }^2=\frac{\sum x_i^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2$
$=\frac{n(n+1)(2n+1)}{6n}-{\left[\frac{n(n+1)}{2n}\right]}^2$
$=\frac{\left(2n^2+3n+1\right)}{6}-{\left(\frac{n+1}{4}\right)}^2$
$=\frac{2\left[2n^2+3n+1\right]-3\left(n^2+1+2n\right)}{12}$
$=\frac{4n^2+6n+2-3n^2-3-6n}{12}=\frac{n^2-1}{12}$