1. **State the problem:** We need to find the value of the sum $$\sum_{n=2}^{25} 200(0.85)^{n-1}$$ and round it to the nearest integer. 2. **Identify the type of sum:** This is a geometric series with the first term at $n=2$. 3. **General formula for geometric series:** $$S = a \frac{1-r^m}{1-r}$$ where $a$ is the first term, $r$ is the common ratio, and $m$ is the number of terms. 4. **Find the first term $a$:** At $n=2$, the term is $$200(0.85)^{2-1} = 200(0.85)^1 = 200 \times 0.85 = 170$$ 5. **Common ratio $r$:** $$r = 0.85$$ 6. **Number of terms $m$:** From $n=2$ to $n=25$ inclusive, number of terms is $$25 - 2 + 1 = 24$$ 7. **Calculate the sum:** $$S = 170 \times \frac{1 - (0.85)^{24}}{1 - 0.85}$$ 8. **Calculate powers and denominator:** $$1 - 0.85 = 0.15$$ Calculate $(0.85)^{24}$: Using a calculator, $(0.85)^{24} \approx 0.03876$ 9. **Substitute values:** $$S = 170 \times \frac{1 - 0.03876}{0.15} = 170 \times \frac{0.96124}{0.15}$$ 10. **Simplify:** $$\frac{0.96124}{0.15} \approx 6.4083$$ 11. **Final sum:** $$S \approx 170 \times 6.4083 = 1089.41$$ 12. **Round to nearest integer:** $$\boxed{1089}$$