Statement-1: The variance of the first n even | vedclass

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Question

(B) The first $n$ even natural numbers are $2, 4, 6, \dots, 2n$.The sum of these numbers is $\sum x_i = 2(1 + 2 + \dots + n) = 2 \cdot \frac{n(n + 1)}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The first $n$ even natural numbers are $2, 4, 6, \dots, 2n$.The sum of these numbers is $\sum x_i = 2(1 + 2 + \dots + n) = 2 \cdot \frac{n(n + 1)}{2} = n(n + 1)$.The mean is $\bar{x} = \frac{n(n + 1)}{n} = n + 1$.The sum of the squares is $\sum x_i^2 = 2^2 + 4^2 + \dots + (2n)^2 = 4(1^2 + 2^2 + \dots + n^2) = 4 \cdot \frac{n(n + 1)(2n + 1)}{6} = \frac{2n(n + 1)(2n + 1)}{3}$.The variance $\sigma^2$ is given by $\frac{\sum x_i^2}{n} - (\bar{x})^2$.$\sigma^2 = \frac{2(n + 1)(2n + 1)}{3} - (n + 1)^2 = (n + 1) \left[ \frac{4n + 2}{3} - (n + 1) ight] = (n + 1) \left[ \frac{4n + 2 - 3n - 3}{3} ight] = (n + 1) \frac{n - 1}{3} = \frac{n^2 - 1}{3}$.Since $\frac{n^2 - 1}{3} 
eq \frac{n^2 - 1}{4}$,Statement-$1$ is false.Statement-$2$ is a standard mathematical result and is true.

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

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