The variance of first n natural numbers is - Infinity Learn by Sri Chaitanya

A
$\frac{n^{2} + 1}{12}$
B
$\frac{n^{2} - 1}{12}$
C
$\frac{(n + 1) (2 n + 1)}{6}$
D
none of these

Solution:

We have,

$\begin{array}{l} σ^{2} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}^{2} - {(\frac{1}{n} \sum_{i = 1}^{n} x_{i})}^{2} \\ \Rightarrow σ^{2} = \frac{1}{n} (1^{2} + 2^{2} + \dots + n^{2}) - {(\frac{1}{n} (1 + 2 + \dots + n))}^{2} \\ \Rightarrow σ^{2} = \frac{1}{n} \times \frac{n (n + 1) (2 n + 1)}{6} - {(\frac{n + 1}{2})}^{2} = \frac{n^{2} - 1}{12} \end{array}$

The variance of first $n$ natural numbers is

Solution:

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