# The S.D of first $n$ naturals is

$\begin {array} {1 1} (A)\;\large\frac{n+1}{2} & \quad (B)\;\sqrt{\large\frac{n / n+1}{2}} \\ (C)\;\sqrt {\large\frac{n^2-1}{12}} & \quad (D)\;None \: of \: these \end {array}$

$\sigma = \sqrt { \large\frac{1}{n} \Sigma x^2 - \bigg( \large\frac{ \Sigma x}{n} \bigg)^2}$
$= \sqrt { \large\frac{n(n+1)(2n+1)}{6n}- \bigg( \large\frac{n(n+1)}{2n} \bigg)^2}$
$= \sqrt{ \large\frac{n+1}{2} \bigg( \large\frac{2n+1}{3}-\large\frac{n+1}{2} \bigg)}$
$= \sqrt{ \large\frac{n+1}{2} \bigg( \large\frac{n-1}{6} \bigg) }$
$= \sqrt { \large\frac{n^2-1}{12}}$
Ans : (C)