Question

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

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

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 $=\frac{1}{n}(\sum x_1{)}^2-(\overline{x}{)}^2=\frac{1}{n}(2^2+4^2+...+(2n{)}^2)-(n+1{)}^2$
$=\frac{1}{n}{2}^2(1^2+2^2+...+n^2)-(n+1{)}^2$
$=\frac{4}{n}\frac{n(n+1)(2n+1)}{6}-(n+1{)}^2$
$=\frac{2}{3}(n+1)(2n+1)-(n+1{)}^2$
$=\frac{n+1}{3}\left[2\right(2n+1)-3(n+1\left)\right]$
$=\frac{(n+1)}{3}(n-1)=\frac{n^2-1}{3}$