Question

Find variance of first 2n natural numbers.

Answer 1

Step 1: Find the Mean of the First 2n Natural Numbers
The mean is the sum of the numbers divided by the total count of numbers: \[ \text{Mean} = \frac{(1 + 2 + 3 + ... + 2n)}{2n} \]

Step 2: Simplify the Mean Using the Arithmetic Series Formula
Using the sum formula for an arithmetic series, we get: \[ \text{Sum} = \frac{2n(1 + 2n)}{2} \] Simplifying the mean, we get: \[ \text{Mean} = \frac{(2n + 1)}{2} \]

Step 3: Define the Variance
Variance is defined as the average of squared differences from the mean: \[ \text{Variance} = \frac{[(1 - \text{mean})^2 + (2 - \text{mean})^2 + ... + (2n - \text{mean})^2]}{2n} \]

Step 4: Simplify the Variance Expression
Expanding and simplifying the expression, we obtain: \[ \text{Variance} = \frac{[4n^3 - 4n]}{12n} \]

Step 5: Further Simplify the Variance
This simplifies to: \[ \text{Variance} = \frac{(4n^2 - 1)}{12} \]

Step 6: Final Result
Therefore, the variance of the first 2n natural numbers is: \[ \text{Variance} = \frac{(4n^2 - 1)}{12} \]