To verify whether $f(x) = \frac{2x}{k(k+1)}$ can serve as the probability distribution of a discrete random variable $x$, we need to check if it satisfies the two conditions for a probability distribution:

1. Non-negativity: $f(x) \geq 0$ for all values of $x$.
2. Normalization: The sum of probabilities must equal 1, i.e., $\sum_{x=1}^{k} f(x) = 1$.

Let's check these conditions one by one.

1. Non-negativity

Since $x \in \{1, 2, 3, \dots, k\}$ and $k(k+1)$ is always positive, the function $f(x) = \frac{2x}{k(k+1)}$ is clearly non-negative for all $x$. Therefore, the non-negativity condition is satisfied.

2. Normalization

Now, we need to check if the sum of all probabilities equals 1: $\sum_{x=1}^{k} f(x) = \sum_{x=1}^{k} \frac{2x}{k(k+1)} = \frac{2}{k(k+1)} \sum_{x=1}^{k} x.$ The sum of the first $k$ positive integers is a well-known result: $\sum_{x=1}^{k} x = \frac{k(k+1)}{2}.$ Substituting this into the previous equation: $\sum_{x=1}^{k} f(x) = \frac{2}{k(k+1)} \cdot \frac{k(k+1)}{2} = 1.$ Thus, the normalization condition is satisfied.

Conclusion

Since both conditions (non-negativity and normalization) are satisfied, the function $f(x) = \frac{2x}{k(k+1)}$ can indeed serve as the probability distribution of a random variable for $x = 1, 2, 3, \dots, k$.

Would you like further clarification on this, or details on related concepts?

Here are 5 related questions:

1. How does this probability distribution behave as $k$ increases?
2. Can we interpret the shape of this probability distribution?
3. How would you calculate the expected value $E(x)$ for this distribution?
4. What happens if you change the form of $f(x)$, for example, to a different polynomial?
5. How does the normalization process differ for continuous distributions?

Tip: The sum of the first $n$ integers is a common result in probability and combinatorics. It's useful to memorize formulas for sums to simplify problems like these.