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 $\frac{n(n+1)(2n+1)}{6}$.

• 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 $=\left[\frac{1}{n}\Sigma {\left(x_i\right)}^2\right]-(\bar{x}{)}^2$
$=\frac{1}{n}\left[2^2+4^2+\dots +(2n{)}^2\right]-(n+1{)}^2$
$=\frac{1}{n}{2}^2\left[1^2+2^2+\dots +n^2\right]-(n+1{)}^2$
$=\frac{4}{n}\frac{n(n+1)(2n+1)}{6}-(n+1{)}^2$
$=\frac{(n+1)[2(2n+1)-3(n+1)]}{3}$
$=\frac{(n+1)(n-1)}{3}=\frac{n^2-1}{3}$
$\therefore$ Statement $I$ is false.