# First n natural numbers:

$\begin{array}{1 1}(A)\;2n\\(B)\;n+1\\(C)\;\large\frac{n^2-1}{12}\\(D)\;0\end{array}$

## 1 Answer

Step 1:
First n natural numbers are $1,2,3,4....n$
$x_i=1;x_i^2=1^2$
$x_i=2;x_i^2=2^2$
$x_i=3;x_i^2=3^2$
$x_4=1;x_i^2=4^2$
.............. $x_i=n;x_i^2=n^2$
Total $= \large\frac{n(n-1)}{2} \qquad \large\frac{n(n-1)(2n+1)}{6}$
Step 2:
Mean $= \large\frac{\sum x_i}{n}$
$\bar{x}=\large\frac{n(n+1)}{2n}$
$\qquad= \large\frac{n+1}{2}$
Step 3:
Variance $=\large\frac{\sum x_i^2}{n} - \bigg(\large\frac{\sum x_i}{n} \bigg)^2$
$\qquad= \large\frac{n(n+1)(2n+1)}{6n)}$
$\qquad= \large\frac{(n+1)(2n+1)}{6} -\large\frac{(n+1)^2}{4}$
$\qquad= \large\frac{(n+1)}{2} \bigg[ \frac{2n+1}{3} -\large\frac{n+1}{2} \bigg]$
$\qquad= \large\frac{(n+1)}{2} \bigg[ \large\frac{4n+2 -3n-3}{6} \bigg]$
$\qquad= \bigg( \large\frac{n+1}{2}\bigg) \bigg[\large\frac{n-1}{6} \bigg]$
$\qquad= \large\frac{n^2-1}{12}$
Hence C is the correct answer.
answered Jun 27, 2014 by

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