1. The problem is to explain how the sum $\sum_{t=1}^4 (1+0.06)^t \times 500$ is simplified to $500 \times \left( \frac{0.06}{1 - (1.06)^{-4}} \right)$. 2. This sum is a geometric series where each term is $500 \times (1.06)^t$ for $t=1$ to $4$. 3. The formula for the sum of a geometric series $S_n = a \times \frac{r^n - 1}{r - 1}$ applies, where $a$ is the first term and $r$ is the common ratio. 4. Here, the first term $a = 500 \times 1.06$ (when $t=1$) and the ratio $r = 1.06$. 5. So, the sum is: $$ S_4 = 500 \times 1.06 \times \frac{(1.06)^4 - 1}{1.06 - 1} $$ 6. Simplify the denominator: $$ 1.06 - 1 = 0.06 $$ 7. Substitute back: $$ S_4 = 500 \times 1.06 \times \frac{(1.06)^4 - 1}{0.06} $$ 8. To express the sum in the form given, multiply numerator and denominator inside the fraction by $(1.06)^{-4}$ to get: $$ \frac{(1.06)^4 - 1}{0.06} = \frac{1 - (1.06)^{-4}}{0.06 \times (1.06)^{-4}} $$ 9. Rearranging gives: $$ S_4 = 500 \times \frac{0.06}{1 - (1.06)^{-4}} $$ 10. This matches the expression: $$ 500 \times \left( \frac{0.06}{1 - (1.06)^{-4}} \right) $$ This shows the equivalence and how the sum formula is manipulated to the given form.