Let's break down the series you provided and find a general expression for it.

The series is: $2 + 2(3) + 2(3^2) + \cdots + 2(3^n)$

This can be written as: $2(1) + 2(3) + 2(3^2) + \cdots + 2(3^n)$

We can factor out the common factor of 2: $2 \left(1 + 3 + 3^2 + \cdots + 3^n\right)$

The expression inside the parentheses is a geometric series with the first term $a = 1$ and the common ratio $r = 3$.

The sum $S$ of the first $n+1$ terms of a geometric series is given by: $S = \frac{a(r^{n+1} - 1)}{r - 1}$

Substituting $a = 1$ and $r = 3$, we get: $S = \frac{1(3^{n+1} - 1)}{3 - 1} = \frac{3^{n+1} - 1}{2}$

So, the sum of the series is: $2 \times \frac{3^{n+1} - 1}{2}$

Simplifying, we get: $3^{n+1} - 1$

Thus, the expression for the sum of the series $2 + 2(3) + 2(3^2) + \cdots + 2(3^n)$ is: $\boxed{3^{n+1} - 1}$

Would you like any more details on this solution?

Here are 5 related questions:

1. How would you find the sum if the common ratio were different from 3?
2. What is the sum of the series if $n = 5$?
3. How does the formula change if the first term $a$ is not 1?
4. Can you express the sum of an infinite geometric series?
5. What happens to the sum as $n$ approaches infinity?

Tip: Remember, geometric series are powerful tools for finding sums where each term is a constant multiple of the previous one.