1. **State the problem:** We need to find the value of the sum $$\sum_{n=0}^{53} 40(1.03)^{n+1}$$ and round it to the nearest integer. 2. **Identify the type of series:** This is a geometric series where each term is of the form $$ar^n$$ but here the exponent is $$n+1$$, so the first term corresponds to $$n=0$$ giving $$40(1.03)^1$$. 3. **Formula for the sum of a geometric series:** For $$n$$ terms starting from $$k=0$$, the sum is $$$S_n = a \frac{r^n - 1}{r - 1}$$$ where $$a$$ is the first term and $$r$$ is the common ratio. 4. **Apply the formula:** - First term $$a = 40(1.03)^1 = 40 \times 1.03 = 41.2$$ - Common ratio $$r = 1.03$$ - Number of terms $$n = 54$$ (since from $$n=0$$ to $$n=53$$ is 54 terms) 5. **Calculate the sum:** $$$S_{54} = 41.2 \times \frac{(1.03)^{54} - 1}{1.03 - 1} = 41.2 \times \frac{(1.03)^{54} - 1}{0.03}$$$ 6. **Evaluate powers and simplify:** Calculate $$ (1.03)^{54} $$ approximately: $$$ (1.03)^{54} \approx 4.801 $$ So, $$$S_{54} = 41.2 \times \frac{4.801 - 1}{0.03} = 41.2 \times \frac{3.801}{0.03} = 41.2 \times 126.7 = 5219.64$$$ 7. **Round the result:** The sum rounded to the nearest integer is $$\boxed{5220}$$. This is the value of the given geometric series sum.