To solve the given problem, let's break it into steps:

Problem Summary:

We are tasked with finding the expected value $E[X]$ of a random variable $X$ that takes on the powers of 2 in the range $[2^a, 2^b]$, with equal probability.

Key Information:

1. $a$ and $b$ are positive integers such that $a \leq b$.
2. $X$ can take the values $2^a, 2^{a+1}, \dots, 2^b$, with equal probability.

Step 1: Determine the values of $X$

The possible values for $X$ are $2^a, 2^{a+1}, \dots, 2^b$, which gives $b-a+1$ values.

Step 2: Probability of each value

Since all values have equal probability, the probability of each value is:

$P(X = 2^k) = \frac{1}{b-a+1}, \quad \text{for } k = a, a+1, \dots, b.$

Step 3: Expected Value Formula

The expected value of $X$ is:

$E[X] = \sum_{k=a}^b P(X = 2^k) \cdot 2^k.$

Substitute $P(X = 2^k) = \frac{1}{b-a+1}$:

$E[X] = \frac{1}{b-a+1} \sum_{k=a}^b 2^k.$

Step 4: Simplify the Summation

The sum $\sum_{k=a}^b 2^k$ is a geometric series with the first term $2^a$ and ratio 2:

$\sum_{k=a}^b 2^k = 2^a (1 + 2 + 2^2 + \cdots + 2^{b-a}).$

This can be simplified using the geometric sum formula:

$\sum_{k=a}^b 2^k = 2^a \left( \frac{2^{b-a+1} - 1}{2 - 1} \right) = 2^a \left( 2^{b-a+1} - 1 \right).$

Step 5: Final Expression for $E[X]$

Substitute back into the formula for $E[X]$:

$E[X] = \frac{1}{b-a+1} \cdot 2^a \left( 2^{b-a+1} - 1 \right).$

Simplify:

$E[X] = \frac{2^{b+1} - 2^a}{b-a+1}.$

Step 6: Match the Answer

The correct option is (b): $\frac{2^{b+1} - 2^a}{b-a+1}$.

Do you want further clarification, or have more questions about this problem?

Related Questions:

1. What is the formula for the sum of a geometric series, and how is it derived?
2. How would the result change if the random variable took values with non-equal probabilities?
3. Can we generalize this result for powers other than 2 (e.g., 3)?
4. What does the expected value tell us about the distribution of the random variable?
5. How would you calculate the variance of this random variable?

Tip:

For random variables defined over a finite range with equal probabilities, the expected value simplifies to the weighted average of the possible outcomes.