The given expression is:

$2 + 2(3) + 2(3^2) + 2(3^3) + \dots + 2(3^n)$

This is a geometric series where each term after the first is multiplied by 3. The first term is 2, and the common ratio $r$ is 3.

The general formula for the sum $S_n$ of a geometric series is:

$S_n = a \frac{1 - r^{n+1}}{1 - r}$

Where:

• $a$ is the first term (in this case, 2),
• $r$ is the common ratio (in this case, 3),
• $n$ is the highest exponent of 3.

However, since each term has a factor of 2, we can factor it out. So, the sum becomes:

$S_n = 2 \left( 1 + 3 + 3^2 + 3^3 + \dots + 3^n \right)$

This is a geometric series with the first term 1 and common ratio 3. Using the geometric sum formula for $1 + 3 + 3^2 + \dots + 3^n$, we have:

$S_n = 2 \cdot \frac{3^{n+1} - 1}{3 - 1}$

Simplifying:

$S_n = 2 \cdot \frac{3^{n+1} - 1}{2}$

So, the expression simplifies to:

$S_n = 3^{n+1} - 1$

Thus, the expression for the sum of the given series is:

$S_n = 3^{n+1} - 1$

Do you want more details or have any questions?

Here are 5 related questions for further exploration:

1. What is the sum of the series when $n = 4$?
2. How does the sum change if the common ratio is 4 instead of 3?
3. How would you modify the formula if the first term is not 2?
4. What is the general form of a geometric series?
5. How do geometric series differ from arithmetic series?

Tip: Always check the first few terms of a series to verify if it follows a geometric progression pattern before applying formulas.