To solve for the value of $c$ in the given probability mass function (pmf) of the discrete random variable $X$, we need to use the fact that the sum of all probabilities in a pmf must equal 1.

Problem Recap:

The pmf is given as: $p_X(2^{-k}) = P(X = 2^{-k}) = \frac{c}{k(k+1)}, \quad k = 2, 3, 4, \ldots$ where $c$ is a positive constant.

Step 1: Write the total probability condition

The total probability must sum to 1: $\sum_{k=2}^{\infty} p_X(2^{-k}) = 1$

Substituting the pmf: $\sum_{k=2}^{\infty} \frac{c}{k(k+1)} = 1$

Step 2: Simplify the summation expression

Notice that: $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$ This is a telescoping series.

So, the sum becomes: $\sum_{k=2}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right)$

The telescoping nature means that most terms will cancel out, leaving: $\left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \cdots$ This simplifies to: $\frac{1}{2}$

Step 3: Solve for $c$

Now, equate the sum to 1: $c \times \frac{1}{2} = 1$ $c = 2$

Final Answer:

The value of $c$ is $\boxed{2}$.

Would you like further details on any step? Here are five related questions to deepen your understanding:

1. How does the telescoping series help in simplifying summations?
2. What would the pmf look like for different values of $k$?
3. How is the constant $c$ determined in other types of discrete distributions?
4. Can you derive the pmf of $X$ if it had a different form for $k$?
5. How would the problem change if the summation range started from a different $k$?

Tip: Always remember that in probability, the total sum of the pmf must equal 1. This is a key concept in solving for unknown constants.