1. **State the problem:** We want to solve the summation $$\sum_{i=0}^{39} (1 + 0.05)^i$$ where the base is $1 + r = 1.05$ and the upper limit is 39. 2. **Formula used:** This is a geometric series with first term $a = (1.05)^0 = 1$ and common ratio $r = 1.05$. The sum of the first $n+1$ terms of a geometric series is given by: $$S_{n} = \frac{a(r^{n+1} - 1)}{r - 1}$$ 3. **Apply the formula:** Here, $n = 39$, $a = 1$, and $r = 1.05$. $$S_{39} = \frac{1(1.05^{40} - 1)}{1.05 - 1}$$ 4. **Calculate the denominator:** $$1.05 - 1 = 0.05$$ 5. **Calculate the numerator:** $$1.05^{40} - 1$$ Using a calculator, $1.05^{40} \approx 7.03999$. So, $$7.03999 - 1 = 6.03999$$ 6. **Put it all together:** $$S_{39} = \frac{6.03999}{0.05}$$ 7. **Simplify the fraction:** $$\frac{6.03999}{0.05} = 6.03999 \times \frac{1}{0.05} = 6.03999 \times 20 = 120.7998$$ **Final answer:** $$\sum_{i=0}^{39} (1.05)^i \approx 120.8$$