Question

The variance of first $n$ natural numbers is

• A. $\frac{n(n+1)}{2}$
• B. $\frac{(n+1)(n+5)}{12}$
• C. $\frac{(n+1)(n-5)}{12}$
• D. $\frac{\left(n^2-1\right)}{12}$

Answer 1

Answer: (D)

Variance of first $n$ natural numbers
$ =\frac{\Sigma n^2}{n}-\left(\frac{\Sigma n}{n}\right)^2$
$ =\frac{n(n+1)(2 n+1)}{6 n}-\left(\frac{n(n+1)}{2 n}\right)^2$
$ =(n+1)\left[\frac{2 n+1}{6}-\frac{(n+1)}{4}\right] $
$ =\frac{(n+1)}{12}[4 n+2-3 n-3]$
$=\frac{(n+1)}{12} \times(n-1)=\frac{n^2-1}{12}$