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

To determine the truth of the given statements, let's analyze them one by one.

Statement-1: The variance of the first \( n \) even natural numbers is \(\frac{n^2 - 1}{4}\).

To find the variance of the first \( n \) even natural numbers, let's consider the sequence: \( 2, 4, 6, \ldots, 2n \).

1. The mean (\(\mu\)) of these numbers is calculated as: \(\mu = \frac{\text{Sum of first } n \text{ even numbers}}{n}\). The sum of the first \( n \) even natural numbers is \( n(n + 1) \). Therefore, the mean is: \(\mu = \frac{n(n+1)}{n} = (n+1)\).
2. The variance is calculated using the formula: \(\text{Variance} = \frac{\sum (x_i - \mu)^2}{n}\).
3. First, compute the sum of squares of the even numbers: \( (2^2 + 4^2 + 6^2 + \ldots + (2n)^2) = 4(1^2 + 2^2 + 3^2 + \ldots + n^2) \). Using the formula for the sum of squares of the first \( n \) natural numbers: \(\frac{n(n + 1)(2n +1)}{6}\), we get: \(\text{Sum of squares} = 4 \cdot \frac{n(n+1)(2n+1)}{6}\).
4. The variance can be calculated as follows: \(\text{Variance} = \frac{4 \cdot \frac{n(n+1)(2n+1)}{6} - n(n+1)^2}{n}\). Upon simplifying, the variance does not reduce to \(\frac{n^2 - 1}{4}\).

So, Statement-1 is false.

Statement-2: The sum of the first \( n \) natural numbers is \(\frac{n(n + 1)}{2}\) and the sum of squares of the first \( n \) natural numbers is \(\frac{n(n + 1)(2n +1)}{6}\).

Both formulas are standard and widely accepted in mathematics:

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

These formulas are correct, hence Statement-2 is true.

Conclusion: According to the above analysis, Statement-1 is false and Statement-2 is true. Therefore, the correct answer is: Statement-1 is false, Statement-2 is true.