Find an expression for 2+2(3)+2(32)+⋯+2(3n)2 + 2(3) + 2(3^2) + \cdots + 2(3^n)2+2(3)+2(32)+⋯+2(3n).

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.

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

Explore how to derive the expression for the sum of the series 2 + 2(3) + 2(3^2) + ⋯ + 2(3^n). This problem covers essential concepts in algebra, including geometric series and their sums. Perfect for students in Grades 9-11.

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