Question

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

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

Answer 1

Answer: (D)

Statement-2 is true
Statement-1:
Sum of n even natural numbers $=n(n+1)$
Mean $\left(\bar{x}\right)=\frac{n\left(n+1\right)}{n}=n+1$
Variance $=\left[\frac{1}{n}\sum {\left(x_i\right)}^2\right]-{\left(\bar{x}\right)}^2=\frac{1}{n}\left[2^2+4^2+.....+{\left(2n\right)}^2\right]-{\left(n+1\right)}^2$
$=\frac{1}{n}{2}^2\left[1^2+2^2+.....+n^2\right]-{\left(n+1\right)}^2=\frac{4}{n}\frac{n\left(n+1\right)\left(2n+1\right)}{6}-{\left(n+1\right)}^2$
$=\frac{\left(n+1\right)\left[2\left(2n+1\right)-3\left(n+1\right)\right]}{3}=\frac{\left(n+1\right)\left[4n+2-3n-3\right]}{3}=\frac{\left(n+1\right)\left(n-1\right)}{3}=\frac{n^2-1}{3}$
$\therefore$ Statement 1 is false.