Question

Variance of first 2n natural numbers?

Answer 1

To find the mean and variance of \( n \) natural numbers:

Step 1: Mean
The sum of the first \( n \) natural numbers is given by the formula: \[ \frac{n(n+1)}{2} \] To calculate the mean, we divide the sum by \( n \). Therefore, the mean of the first \( n \) natural numbers is: \[ \frac{n(n+1)}{2n} = \frac{n+1}{2} \] Thus, the mean is \( \frac{n+1}{2} \).

Step 2: Variance
The variance is a measure of how much the numbers deviate from the mean. It can be calculated using the formula: \[ \text{Variance} = \frac{1}{n} \sum_{i=1}^n (x_i - \mu)^2 \] However, an alternative formula for the variance of the first \( n \) natural numbers is given by: \[ \frac{6n(n+1)(2n+1) - (n+1)^2}{(2n+1)} \] Simplifying this expression gives: \[ \frac{12n^2 - 1}{(2n+1)} \]

Final Answer:
Hence, the correct option for the variance is: \[ \frac{12n^2 - 1}{(2n+1)} \]