Question

Find variance of first 2n natural numbers.

Answer 1

To find the variance of the first 2n natural numbers, we first need to find the mean of the sequence.

The mean is simply the sum of the numbers divided by the total count of numbers:
mean = $\frac{(1 + 2 + 3 + ... + 2n)}{2n}$

Using the formula for the sum of an arithmetic series, we can simplify the expression for the mean to:
mean = $\frac{(2n + 1)}{2}$
Next, we need to find the variance.

The variance is defined as the average of the squared differences from the mean. In other words:
variance = $\frac{[(1 - mean)^2 + (2 - mean)^2 + ... + (2n - mean)^2]}{2n}$

We can simplify this expression by expanding the squares and using the formula for the sum of the first n natural numbers:
variance = $\frac{[n(2n + 1)(4n + 1) - 6n(2n + 1) + 6n^2]}{(12n)}$

Simplifying this expression, we get:
variance = $\frac{(4n^2 - 1)}{12}$

Therefore, the variance of the first 2n natural numbers is $\frac{(4n^2 - 1)}{12}$.