Question

The variance of first n even natural numbers is $\frac{n^2 - 1}{4}$. 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-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

The Correct Option is C

The correct answer is C:Statement-1 is false, Statement-2 is true
Given that;
Sum of first ‘n’ even natural numbers 
\(=2+4+6+.....+2n\)
\(=2(1+2+.....n)\)
\(\frac{2(n+1)n}{2}=n(n+1)\)
For the numbers 2, 4, 6, 8, ......., 2n 
\(\bar{x} = \frac{2\left[n\left(n+1\right)\right]}{2n} =\left(n+1\right)\) 
And \(Var = \frac{\sum\left(x -\bar{x}\right)^{2}}{2n} = \frac{\sum x^{2}}{n} - \left(\bar{x}\right)^{2}\)
\(= \frac{4\sum n^{2}}{n} - \left(n+1\right)^{2}= \frac{4n\left(n+1\right)\left(2n+1\right)}{6n} - \left(n+1\right)^{2}\)
\(= \frac{2\left(2n+1\right)\left(n+1\right)}{3} - \left(n+1\right)^{2}= \left(n+1\right) \left[\frac{4n+2-3n-3}{3} \right]\)
\(= \frac{\left(n+1\right)\left(n-1\right)}{3} = \frac{n^{2}-1}{3}\)
\(\therefore\) Statement-1 is false. Clearly, statement - 2 is true .