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^{2} - 1}{4} and the sum of the squares of first n natural member is \frac{n \left(\right. n + 1 \left.\right) \left(\right. 2 n + 1 \left.\right)}{6}.

Engineering

Mathematics

Measure of Central Tendency

Measure of Dispersion

Question

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 $\frac{n^{2} - 1}{4}$ and the sum of the squares of first n natural member is $\frac{n (n + 1) (2 n + 1)}{6}$ .

Statement-1 is false, statement-2 is true.

Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

Statement-1 is true, statement-2 is false

Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

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Sum of first n even natural numbers is n(n + 1)
Mean $(\bar{x}) = \frac{n (n + 1)}{n} = n + 1$
Variance $\frac{1}{n} \sum {(x_{i})}^{2} - {(\bar{x})}^{2} = \frac{1}{n} (2^{2} + 4^{2} + 6^{2} + \dots \dots + {(2 n)}^{2})$ (n + 1)²
= $\frac{4}{n} (\frac{n (n + 1) (2 n + 1)}{6})$ – (n + 1)² $= \frac{(n + 1) (n - 1)}{3} = \frac{n^{2} - 1}{3}$