The variance of the first 50 even natural numbers is:

We have to find the variance of the first 50 even natural numbers, that is, the numbers

$$2,\,4,\,6,\dots,100.$$

First, let us determine their mean (arithmetic average). These numbers form an arithmetic progression with first term $$a_1=2$$, common difference $$d=2$$ and the 50th term $$a_{50}=100$$. For an arithmetic progression, the mean of all terms equals the mean of the first and last term, so

$$\mu=\frac{a_1+a_{50}}{2}=\frac{2+100}{2}=51.$$

Next, we recall the formula for variance of a finite set of numbers treated as the entire population (not a sample):

$$\sigma^2=\frac{\displaystyle\sum_{i=1}^{n}x_i^2}{n}-\mu^2,$$

where $$n$$ is the number of observations, $$x_i$$ are the individual observations and $$\mu$$ is their mean. Thus, we must compute $$\sum x_i^2$$ for our list.

Each even number in the list can be written as $$x_i=2i$$ with $$i=1,2,\dots,50$$. Therefore,

$$\sum_{i=1}^{50}x_i^2=\sum_{i=1}^{50}(2i)^2=\sum_{i=1}^{50}4i^2=4\sum_{i=1}^{50}i^2.$$

Now we state the standard summation formula for the squares of the first $$n$$ natural numbers:

$$\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}.$$

Substituting $$n=50$$ we get

$$\sum_{i=1}^{50}i^2=\frac{50\,(50+1)\,(2\cdot50+1)}{6}=\frac{50\cdot51\cdot101}{6}.$$

We multiply step by step:

$$50\cdot51=2550,$$ $$2550\cdot101=257{,}550,$$ $$\frac{257{,}550}{6}=42{,}925.$$

Hence,

$$\sum_{i=1}^{50}i^2=42{,}925,$$ $$\sum_{i=1}^{50}x_i^2=4\cdot42{,}925=171{,}700.$$

We now divide by the number of terms $$n=50$$ to get the mean of the squares:

$$\frac{\sum x_i^2}{n}=\frac{171{,}700}{50}=3{,}434.$$

Finally, we substitute into the variance formula:

$$\sigma^2=3{,}434-\mu^2=3{,}434-51^2=3{,}434-2{,}601=833.$$

Hence, the correct answer is Option 4.