Question

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

• A. Statement I is true; Statement II is true; Statement II is a correct explanation for Statement I
• B. Statement I is true; Statement II is true; Statement II is not a correct explanation for Statement I
• C. Statement I is true; Statement II is false
• D. Statement I is false; Statement II is true

Answer 1

Answer: (D)

Statement II is true.
I. Sum of $n$ even natural numbers $=n(n+1)$
Mean $(\bar{x}) =\frac{n(n+1)}{n}=n+1$
Variance $ =\left[\frac{1}{n} \Sigma\left(x_i\right)^2\right]-(\bar{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{(n+1)[2(2 n+1)-3(n+1)]}{3}$
$=\frac{(n+1)(n-1)}{3}=\frac{n^2-1}{3}$
$\therefore$ Statement $I$ is false.