Statement-1: The variance of first n even natural numbers is $\frac{n^{2} - 1}{4} .$
Statement-2: The sum of first n natural numbers is $\frac{n (n + 1)}{2}$ and the sum of square of first n natural numbers is $\frac{n (n + 1) (2 n + 1)}{6} .$

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Statement-1 is true, Statement-2 is true. Statement-2 is not a correct explanation for Statement-1

Statement-1 is false, Statement-2 is true

Statement-1 is true, Statement-2 is false

Statement-1 is true, Statement-2 is true. Statement-2 is a correct explanation for Statement-1

answer is C.

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Detailed Solution

$σ^{2} + μ^{2} = \sum \frac{{x_{i}}^{2}}{n}$

It implies that

$\begin{array}{rcl} σ^{2} & = & \frac{(n + 1) (2 n + 1)}{6} - {(\frac{n + 1}{2})}^{2} \\ = & \frac{n + 1}{2} (\frac{2 n + 1}{3} - \frac{n + 1}{2}) \\ = & \frac{n + 1}{2} (\frac{4 n + 2 - 3 n - 3}{6}) \\ = & \frac{n + 1}{2} \cdot \frac{n - 1}{6} \\ = & \frac{n^{2} - 1}{12} \end{array}$