https://file.aiquickdraw.com/mathbot/file/172676904228545a0uxfg.jpg

The image shows the following mathematical expression:

\[
3^1 + 3^2 + 3^3 + \dots + 3^n = \frac{3^{n+1} - 3}{2}, \quad n \geq 1
\]

This is a formula representing the sum of a geometric series with terms that are powers of 3. The general form of a geometric series is:

\[
S_n = a + ar + ar^2 + \dots + ar^{n-1}
\]

where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. In this case:
- The first term \( a = 3^1 = 3 \),
- The common ratio \( r = 3 \),
- And the series continues until \( 3^n \).

Using the formula for the sum of a geometric series:

\[
S_n = a \frac{r^n - 1}{r - 1}
\]

Here, substituting \( a = 3 \) and \( r = 3 \), we can derive the expression on the right-hand side:

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

This confirms that the formula given in the image is correct.

Would you like more details or have any further questions?

Here are some related questions to explore:
1. How do you derive the formula for the sum of a geometric series in general?
2. What happens to this formula if \( n \to \infty \)?
3. How would the series change if the common ratio was not 3 but another number?
4. Can you generalize this formula for different starting values of \( a \)?
5. How can you use this formula to solve practical problems involving geometric growth?

**Tip**: When working with geometric series, always check if the common ratio is less than or greater than 1 to determine convergence or divergence.

3^1 + 3^2 + 3^3 + ... + 3^n = (3^(n+1) - 3) / 2, n >= 1

Explore the sum of powers of 3 using a geometric series formula: 3^1 + 3^2 + 3^3 + ... + 3^n = (3^(n+1) - 3) / 2. This explanation is suitable for students in grades 10-12 and involves exponents and series summation.

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