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.