A bag consists of tokens whose labels are selected from all of the first $n$ even natural numbers. The frequency of each label equals half the label itself. Then the variance of all the labels the bag is

The first $n$ even natural numbers are: $2, 4, 6, 8, ..., 2n$

The frequency of each label equals half the label itself, so:

Label	Frequency
2	1
4	2
6	3
$2k$	$k$
$2n$	$n$

Total number of tokens = $1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$

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Finding the Mean ($\mu$):

Mean = $\frac{\text{Sum of all values}}{\text{Total frequency}}$

Sum of all values = $(2 \times 1) + (4 \times 2) + (6 \times 3) + ... + (2n \times n)$

$= \sum_{k=1}^{n} 2k \times k = 2\sum_{k=1}^{n} k^2$

Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$:

Sum = $2 \times \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(2n+1)}{3}$

Therefore: $\mu = \frac{n(n+1)(2n+1)}{3} \times \frac{2}{n(n+1)} = \frac{2(2n+1)}{3}$

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Finding $E[X^2]$:

$E[X^2] = \frac{\text{Sum of (value}^2 \times \text{frequency)}}{\text{Total frequency}}$

Sum of (value$^2 \times$ frequency) = $(2^2 \times 1) + (4^2 \times 2) + (6^2 \times 3) + ... + ((2n)^2 \times n)$

$= \sum_{k=1}^{n} (2k)^2 \times k = 4\sum_{k=1}^{n} k^3$

Using the formula $\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2$:

Sum = $4 \times \frac{n^2(n+1)^2}{4} = n^2(n+1)^2$

Therefore: $E[X^2] = n^2(n+1)^2 \times \frac{2}{n(n+1)} = 2n(n+1)$

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Finding Variance:

Variance = $E[X^2] - \mu^2$

$= 2n(n+1) - \left[\frac{2(2n+1)}{3}\right]^2$

$= 2n(n+1) - \frac{4(2n+1)^2}{9}$

$= \frac{18n(n+1) - 4(2n+1)^2}{9}$

Expanding $(2n+1)^2 = 4n^2 + 4n + 1$:

$= \frac{18n^2 + 18n - 4(4n^2 + 4n + 1)}{9}$

$= \frac{18n^2 + 18n - 16n^2 - 16n - 4}{9}$

$= \frac{2n^2 + 2n - 4}{9}$

The answer is Option 2.