1. **Problem statement:** (c) Write an expression for $A_n$ in terms of $x$ on the day before Mary turned 18, showing clearly the value of $n$. 2. **Understanding the problem:** Mary makes monthly deposits of $x$ into an account earning interest. We want the amount $A_n$ just before she turns 18. 3. **Formula used:** For an account with monthly deposits $x$, monthly interest rate $i$, and $n$ months, the amount is given by the future value of an annuity: $$A_n = x \times \frac{(1+i)^n - 1}{i}$$ 4. **Value of $n$:** Since Mary deposits monthly until just before she turns 18, and assuming she started at birth, $n = 18 \times 12 = 216$ months. 5. **Expression for $A_n$:** $$A_{216} = x \times \frac{(1+i)^{216} - 1}{i}$$ --- (d) Mary's grandparents want $A_{216} \geq 20000$. Given $i=0.4\% = 0.004$, find minimum $x$: $$20000 \leq x \times \frac{(1+0.004)^{216} - 1}{0.004}$$ Calculate: $$ (1.004)^{216} = e^{216 \times \ln(1.004)} \approx e^{216 \times 0.003992} = e^{0.861} \approx 2.365 $$ So: $$ \frac{2.365 - 1}{0.004} = \frac{1.365}{0.004} = 341.25 $$ Then: $$ 20000 \leq x \times 341.25 \implies x \geq \frac{20000}{341.25} \approx 58.62 $$ Minimum monthly deposit $x$ is approximately 59. --- (e) Mary invests 15000 at 0.4% monthly interest, withdraws 1000 yearly on birthday. Find time until account is empty. 1. Interest rate per month: $i=0.004$. 2. Withdrawals: $1000$ every 12 months. 3. Model as a discrete process with balance $B_k$ just after $k$ withdrawals: $$B_0 = 15000$$ After 12 months, balance grows: $$B_0 (1+i)^{12}$$ Then withdraw 1000: $$B_1 = B_0 (1+i)^{12} - 1000$$ Similarly: $$B_k = B_{k-1} (1+i)^{12} - 1000$$ 4. This is a linear recurrence: $$B_k = B_0 (1+i)^{12k} - 1000 \sum_{j=0}^{k-1} (1+i)^{12j}$$ Sum of geometric series: $$\sum_{j=0}^{k-1} (1+i)^{12j} = \frac{(1+i)^{12k} - 1}{(1+i)^{12} - 1}$$ 5. So: $$B_k = (1+i)^{12k} \left[ B_0 - \frac{1000}{(1+i)^{12} - 1} \right] + \frac{1000}{(1+i)^{12} - 1}$$ 6. Find $k$ when $B_k \leq 0$: Calculate: $$r = (1+i)^{12} = (1.004)^{12} = e^{12 \times 0.003992} = e^{0.0479} \approx 1.049$$ Then: $$B_k = r^k \left[15000 - \frac{1000}{r - 1} \right] + \frac{1000}{r - 1}$$ Calculate denominator: $$r - 1 = 0.049$$ $$\frac{1000}{0.049} \approx 20408.16$$ Inside bracket: $$15000 - 20408.16 = -5408.16$$ So: $$B_k = r^k (-5408.16) + 20408.16$$ Set $B_k = 0$: $$0 = r^k (-5408.16) + 20408.16$$ $$r^k = \frac{20408.16}{5408.16} \approx 3.774$$ Take natural log: $$k \ln(r) = \ln(3.774)$$ $$k = \frac{\ln(3.774)}{\ln(1.049)} = \frac{1.329}{0.0479} \approx 27.75$$ 7. So after about 28 withdrawals (years), the account is empty. **Final answers:** (c) $$A_{216} = x \times \frac{(1+i)^{216} - 1}{i}$$ with $n=216$. (d) Minimum monthly deposit $x \approx 59$. (e) Account empties after approximately 28 years.