Explanation:

To find the sum of the geometric series \(3 + 3^2 + 3^3 + \ldots + 3^8\), use the formula for the sum of a geometric series:

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

- Here, \(a = 3\) and \(r = 3\).

- The last term is \(3^8\), so there are 8 terms from \(3^1\) to \(3^8\).

- Apply the formula: \( S_8 = 3 \frac{3^8 - 1}{3 - 1} \).

- Calculate \(3^8 = 6561\).

- Substitute in the numbers:

$$ S_8 = 3 \frac{6561 - 1}{2} = 3 \times 3280 = 9840 $$

Therefore, the sum is 9840.

- Option 1: 6561 - This is \(3^8\) alone, not the sum.

- Option 2: 6560 - Incorrect for sum of series.

- Option 3: 9840 - This is the correct sum of the series.

- Option 4: 3280 - A miscalculation involving only \( \frac{6560}{2} \).