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 $\frac{n(n+1)}{2}$ and The sum of squares first n natural numbers is $\frac{n(n+1)(2n+1)}{6}$

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

Statement-1 is false , Statement-2 is true

Sum of first n even natural numbers

$=2+4+6+...+2n=2(1+2+...+n)$

$=2\frac{n(n+1)}{2}=n(n+1)$

Mean $\left(\overline{x}\right)=\frac{n(n+1)}{n}=n+1$

variance = $=\frac{1}{n}\sqrt{\sum (x_1-\overline{x}{)}^2}$

$\sum (x_1-\overline{x}{)}^2=(x_1-\overline{x}{)}^2+(x_2-{\overline{x}}^2+..(x_n-\overline{x}{)}^2$

$\sum (x_1-\overline{x}{)}^2=x_1^2+{\overline{x}}^2-2x_1\overline{x}+..x_n^2+{\overline{x}}^2-2x_n\overline{x}$

In the above term, $\sum 2\overline{x}{x}_i=-2n{\overline{x}}^2$

$=\frac{1}{n}(\sum x_1{)}^2-(\overline{x}{)}^2=\frac{1}{n}(2^2+4^2+...+(2n{)}^2)-(n+1{)}^2$

$=\frac{1}{n}{2}^2(1^2+2^2+...+n^2)-(n+1{)}^2$

$=\frac{4}{n}\frac{n(n+1)(2n+1)}{6}(n+1{)}^2$

$=\frac{2}{3}(n+1)(2n+1)-(n+1{)}^2$

$=\frac{n+1}{3}\left[2\right(2n+1)-3(n+1\left)\right]$

$=\frac{(n+1)}{3}(n-1)=\frac{n^2-1}{3}$