1. **State the problem:** Find the sum of the first 35 terms of the geometric series: $$8, -12, 18, \ldots$$ 2. **Identify the first term and common ratio:** The first term is $$a = 8$$. The common ratio $$r$$ is found by dividing the second term by the first term: $$r = \frac{-12}{8} = -\frac{3}{2}$$. 3. **Formula for the sum of the first $$n$$ terms of a geometric series:** $$S_n = a \frac{1 - r^n}{1 - r}$$ This formula applies when $$r \neq 1$$. 4. **Apply the formula:** Here, $$n = 35$$, $$a = 8$$, and $$r = -\frac{3}{2}$$. Calculate $$r^{35}$$: $$r^{35} = \left(-\frac{3}{2}\right)^{35}$$ (a very large number in magnitude, alternating sign). 5. **Calculate the sum:** $$S_{35} = 8 \times \frac{1 - \left(-\frac{3}{2}\right)^{35}}{1 - \left(-\frac{3}{2}\right)} = 8 \times \frac{1 - \left(-\frac{3}{2}\right)^{35}}{1 + \frac{3}{2}} = 8 \times \frac{1 - \left(-\frac{3}{2}\right)^{35}}{\frac{5}{2}} = 8 \times \frac{2}{5} \times \left(1 - \left(-\frac{3}{2}\right)^{35}\right) = \frac{16}{5} \left(1 - \left(-\frac{3}{2}\right)^{35}\right)$$ 6. **Interpretation:** The sum is $$\frac{16}{5} \left(1 - \left(-\frac{3}{2}\right)^{35}\right)$$. Since $$\left(-\frac{3}{2}\right)^{35}$$ is a very large number, the sum will be dominated by this term. **Final answer:** $$\boxed{S_{35} = \frac{16}{5} \left(1 - \left(-\frac{3}{2}\right)^{35}\right)}$$