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 $\frac{n(n+1)}{2}$ and the sum of squares of first $n$ natural numbers is $\frac{n(n+1)(2n+1)}{6}$

• A. Statement 1 is true, Statement2 is true,Statement 2 is a correct explanation for Statement 1
• B. Statement 1 is true, Statement2 is true;Statement2 is not a correct explanation for statement 1
• C. Statement 1 is true, Statement 2 is false.
• D. Statement 1 is false, Statement 2 is true

Answer 1

The correct option is D Statement 1 is false, Statement 2 is true

Sum of first n even natural numbers

$=2+4+6+...+2n=2(1+2+...+n)$

$=2\frac{n(n+1)}{2}=n(n+1)$

Mean $\left(\overline{x}\right)=\frac{n(n+1)}{n}=n+1$

variance $=\frac{1}{n}(\sum x_1{)}^2-(\overline{x}{)}^2=\frac{1}{n}(2^2+4^2+...+(2n{)}^2)-(n+1{)}^2$

$=\frac{1}{n}{2}^2(1^2+2^2+...+n^2)-(n+1{)}^2$

$=\frac{4}{n}\frac{n(n+1)(2n+1)}{6}-(n+1{)}^2$

$=\frac{2}{3}(n+1)(2n+1)-(n+1{)}^2$

$=\frac{n+1}{3}\left[2\right(2n+1)-3(n+1\left)\right]$

$=\frac{(n+1)}{3}(n-1)=\frac{n^2-1}{3}$

Hence statement $1$ is false while statement $2$ is true