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The given expression is:

\[
2 + 2(3) + 2(3^2) + \cdots + 2(3^n)
\]

This is a geometric series where each term is a multiple of powers of 3. Specifically, the terms are \( 2(3^k) \) for \( k = 0 \) to \( n \).

To simplify this, we can factor out the constant 2 from all terms:

\[
2(1 + 3 + 3^2 + \cdots + 3^n)
\]

The expression inside the parentheses is a geometric series with the first term \( a = 1 \) and the common ratio \( r = 3 \), and it has \( (n+1) \) terms. The sum of a geometric series is given by:

\[
S = \frac{a(r^{n+1} - 1)}{r - 1}
\]

Substituting \( a = 1 \) and \( r = 3 \):

\[
S = \frac{1(3^{n+1} - 1)}{3 - 1} = \frac{3^{n+1} - 1}{2}
\]

Thus, the original expression becomes:

\[
2 \times \frac{3^{n+1} - 1}{2} = 3^{n+1} - 1
\]

So, the simplified expression is:

\[
3^{n+1} - 1
\]

Would you like more details or have any questions?

Here are 5 related questions you might consider:
1. How do you derive the formula for the sum of a geometric series?
2. What happens if the common ratio is less than 1?
3. How would the expression change if the common ratio was different from 3?
4. Can you apply this to other geometric series with different constants?
5. What are some real-world applications of geometric series?

**Tip**: Always check if you can factor out constants before applying formulas for series; it often simplifies the calculations.

Find an expression for 2 + 2(3) + 2(3^2) + ⋯ + 2(3^n).

Learn how to find the expression for 2 + 2(3) + 2(3^2) + ⋯ + 2(3^n) by simplifying a geometric series. This problem demonstrates key concepts from algebra and geometric series formulas, suitable for high school students.

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