1. The problem asks to rewrite the expression $\sum_{i=0}^{39} H(1 + 0.05)^{40 - i}$ in summation form. 2. The given summation is already in summation form: it sums the terms $H(1 + 0.05)^{40 - i}$ from $i=0$ to $i=39$. 3. To clarify, the summation is: $$\sum_{i=0}^{39} H(1.05)^{40 - i}$$ 4. We can rewrite the exponent as $40 - i = (40) - i$, so the terms decrease in exponent as $i$ increases. 5. Alternatively, we can change the index to simplify the expression. Let $j = 40 - i$. When $i=0$, $j=40$, and when $i=39$, $j=1$. 6. Reversing the summation index, the summation becomes: $$\sum_{j=1}^{40} H(1.05)^j$$ 7. This is a geometric series with first term $a = H(1.05)^1 = 1.05H$ and common ratio $r = 1.05$, summed from $j=1$ to $j=40$. 8. The summation in closed form is: $$S = H \sum_{j=1}^{40} (1.05)^j = H \cdot 1.05 \cdot \frac{(1.05)^{40} - 1}{1.05 - 1}$$ 9. This formula uses the geometric series sum rule: $$\sum_{k=1}^n r^k = r \frac{r^n - 1}{r - 1}$$ 10. Therefore, the summation rewritten in summation form with index $j$ is: $$\sum_{j=1}^{40} H(1.05)^j$$ Final answer: $$\sum_{i=0}^{39} H(1.05)^{40 - i} = \sum_{j=1}^{40} H(1.05)^j$$