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| rev | Admin | (Created page with "'''Solution: E''' Let n index the payment times in months. Then for a payment at time n, the discount factor is <math display = "block">\frac{1}{(1.045)^{n / 6}}.</math> The end-of-month payment at month <math>n</math> is be <math>500+(n-1) X</math>. Therefore, we have <math display = "block">30,000=\sum_{n=1}^{60} \frac{500+(n-1) X}{(1.045)^{n / 6}}.</math> {{soacopyright | 2023 }}") | Nov 19'23 at 13:57 | +388 |