Q. The variance of first $n$ numbers is
A) $\frac{n^{2} + 1}{12}$
B) $\frac{n^{2} - 1}{12}$
C) $\frac{(n + 1\ )(2n + 1)}{6}$
D) $\left\lbrack \frac{n(n + 1)}{2} \right\rbrack^{2}$
Explanation
(b)
$\overline{x} = \frac{1 + 2 + 3 + \ldots + n}{n} = \frac{(n + 1)}{2}$
$\therefore\sigma^{2} = \frac{\Sigma\left( x_{i} \right)^{2}}{n} - \left( \overline{x} \right)^{2}$
$= \frac{\Sigma n^{2}}{n} - \left( \frac{n + 1}{2} \right)^{2}$
$= \frac{n(n + 1)(2n + 1)}{6n} - \left( \frac{n + 1}{2} \right)^{2} = \frac{n^{2} - 1}{12}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.