Question

Statement 1: The variance of first n odd natural numbers is $\frac{n^2-1}{3}$.
Statement 2: The sum of first n odd natural number is $n^2$ and the sum of square of first n odd natural numbers is $\frac{n(4n^2+1)}{3}$ .

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

Answer 1

Answer: (C)

Sum of first n even natural numbers
$<br/>=2+4+6+\dots +2n=2(1+2+\dots +n)<br/>$
$<br/>=2\frac{n(n+1)}{2}=n(n+1)<br/>$
$\mathrm{Mean}(\overline{x})=\frac{n(n+1)}{n}=n+1$
$<br/>\text{ variance }=\frac{1}{n}{\left(\sum x_1\right)}^2-(\overline{x}{)}^2=\frac{1}{n}\left(2^2+4^2+\dots +(2n{)}^2\right)-(n+1{)}^2<br/>$
$<br/>=\frac{1}{n}{2}^2\left(1^2+2^2+\dots +n^2\right)-(n+1{)}^2<br/>$
$<br/>=\frac{4}{n}\frac{n(n+1)(2n+1)}{6}-(n+1{)}^2<br/>$
$<br/>=\frac{2}{3}(n+1)(2n+1)-(n+1{)}^2<br/>$
$<br/>=\frac{n+1}{3}[2(2n+1)-3(n+1)]<br/>$
$=\frac{(n+1)}{3}(n-1)=\frac{n^2-1}{3}$
Hence statement 1 is false while statement 2 is true