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
$
=2+4+6+\ldots+2 n=2(1+2+\ldots+n)
$
$
=2 \frac{n(n+1)}{2}=n(n+1)
$
$\operatorname{Mean}(\overline{ x })=\frac{ n ( n +1)}{ n }= n +1$
$
\text { variance }=\frac{1}{ n }\left(\sum x _{1}\right)^{2}-(\overline{ x })^{2}=\frac{1}{ n }\left(2^{2}+4^{2}+\ldots+(2 n )^{2}\right)-( n +1)^{2}
$
$
=\frac{1}{n} 2^{2}\left(1^{2}+2^{2}+\ldots+n^{2}\right)-(n+1)^{2}
$
$
=\frac{4}{n} \frac{n(n+1)(2 n+1)}{6}-(n+1)^{2}
$
$
=\frac{2}{3}(n+1)(2 n+1)-(n+1)^{2}
$
$
=\frac{n+1}{3}[2(2 n+1)-3(n+1)]
$
$ =\frac{(n+1)}{3}(n-1)=\frac{n^{2}-1}{3} $
Hence statement 1 is false while statement 2 is true