Question 1177293
Let's solve this problem step-by-step.

**1. Define the Expected Value**

The expected value of a discrete random variable X is defined as:

* E(X) = Σ [x * P(X = x)]

In our case, X takes values n, and P(X = n) = 2^(n-1) / 3^n. So:

* E(X) = Σ [n * (2^(n-1) / 3^n)] for n = 1, 2, 3, ...

**2. Simplify the Expression**

* E(X) = Σ [n * (2^(n-1) / 3^n)]
* E(X) = Σ [n * (1/2) * (2^n / 3^n)]
* E(X) = (1/2) * Σ [n * (2/3)^n]

**3. Recognize the Series**

We have the series:

* Σ [n * (2/3)^n]

Let's use the formula for the sum of the series Σ nx^(n-1) = 1/(1-x)^2

We can rewrite the sum as:

* Σ [n * (2/3)^n] = (2/3) * Σ [n * (2/3)^(n-1)]

Now, let x = 2/3. Then:

* Σ [n * (2/3)^(n-1)] = 1 / (1 - 2/3)^2
* Σ [n * (2/3)^(n-1)] = 1 / (1/3)^2
* Σ [n * (2/3)^(n-1)] = 1 / (1/9)
* Σ [n * (2/3)^(n-1)] = 9

Therefore:

* Σ [n * (2/3)^n] = (2/3) * 9 = 6

**4. Calculate E(X)**

* E(X) = (1/2) * Σ [n * (2/3)^n]
* E(X) = (1/2) * 6
* E(X) = 3

**Therefore, E(X) = 3.**