1. **Problem statement:** Mary’s grandparents invest $x on 1st January 1998 and continue depositing $x monthly on the first day of each month. The account pays 0.4% interest per month, compounded monthly on the last day. Let $A_n$ be the amount on the last day of the $n$th month after interest is added. 2. **Formula and rules:** The interest rate per month is $r = 0.004$. Each month, the amount grows by a factor of $1.004$ after interest is added, and a deposit of $x$ is made at the start of the month. 3. **(a) Find $A_1$ and show $A_2 = 1.004^2 x + 1.004 x$:** - At the end of month 1, the initial deposit $x$ earns interest: $$A_1 = x \times 1.004$$ - For month 2, the amount after interest on $A_1$ plus the new deposit $x$ grows: $$A_2 = (A_1 + x) \times 1.004 = (1.004 x + x) \times 1.004 = 1.004^2 x + 1.004 x$$ 4. **(b)(i) Expressions for $A_3$ and $A_4$:** - Similarly, $$A_3 = (A_2 + x) \times 1.004 = (1.004^2 x + 1.004 x + x) \times 1.004 = 1.004^3 x + 1.004^2 x + 1.004 x$$ - And, $$A_4 = (A_3 + x) \times 1.004 = (1.004^3 x + 1.004^2 x + 1.004 x + x) \times 1.004 = 1.004^4 x + 1.004^3 x + 1.004^2 x + 1.004 x$$ 5. **(b)(ii) Show amount before 10th birthday:** - The deposits form a geometric series with ratio $1.004$: $$A_n = x(1.004^n + 1.004^{n-1} + \cdots + 1.004) = x \sum_{k=1}^n 1.004^k$$ - Using the geometric series sum formula: $$\sum_{k=1}^n r^k = r \frac{r^n - 1}{r - 1}$$ - Substitute $r=1.004$: $$A_n = x \times 1.004 \frac{1.004^n - 1}{1.004 - 1} = x \times 1.004 \frac{1.004^n - 1}{0.004} = 251 (1.004^n - 1) x$$ - Mary turns 10 years old after 120 months ($n=120$), so: $$A_{120} = 251 (1.004^{120} - 1) x$$ 6. **(c) Expression for $A_n$ before 18th birthday:** - At 18 years, $n = 18 \times 12 = 216$ months. - So, $$A_{216} = 251 (1.004^{216} - 1) x$$ 7. **(d) Minimum $x$ for $A_{216} \geq 20000$:** - Set up inequality: $$251 (1.004^{216} - 1) x \geq 20000$$ - Calculate $1.004^{216}$: $$1.004^{216} = e^{216 \ln(1.004)} \approx e^{216 \times 0.003992} = e^{0.861} \approx 2.365$$ - Substitute: $$251 (2.365 - 1) x \geq 20000$$ $$251 \times 1.365 x \geq 20000$$ $$342.615 x \geq 20000$$ $$x \geq \frac{20000}{342.615} \approx 58.4$$ - Minimum monthly deposit: $$x = 59$$ (nearest dollar) 8. **(e) Withdrawals from $15000$ at 0.4% monthly interest, withdrawing $1000$ yearly:** - Interest rate per month: $r=0.004$. - Withdraw $1000$ every 12 months. - Model as a discrete process with yearly withdrawals: - After 12 months, amount grows by $1.004^{12}$, then withdraw $1000$. - Let $B_k$ be amount after $k$ years just after withdrawal: $$B_{k} = B_{k-1} \times 1.004^{12} - 1000$$ - Initial amount: $$B_0 = 15000$$ - This is an arithmetic-geometric sequence. - Solve for $k$ when $B_k \leq 0$: $$B_k = 15000 \times (1.004^{12})^k - 1000 \sum_{j=0}^{k-1} (1.004^{12})^j$$ - Sum of geometric series: $$\sum_{j=0}^{k-1} r^j = \frac{r^k - 1}{r - 1}$$ - Substitute $r = 1.004^{12} \approx 1.049$: $$B_k = 15000 \times 1.049^k - 1000 \times \frac{1.049^k - 1}{1.049 - 1} = 15000 \times 1.049^k - 1000 \times \frac{1.049^k - 1}{0.049}$$ - Simplify: $$B_k = 15000 \times 1.049^k - 20408.16 (1.049^k - 1) = (15000 - 20408.16) 1.049^k + 20408.16$$ $$B_k = -5408.16 \times 1.049^k + 20408.16$$ - Set $B_k = 0$: $$-5408.16 \times 1.049^k + 20408.16 = 0$$ $$1.049^k = \frac{20408.16}{5408.16} \approx 3.774$$ - Take natural log: $$k \ln(1.049) = \ln(3.774)$$ $$k = \frac{\ln(3.774)}{\ln(1.049)} \approx \frac{1.329}{0.048} = 27.7$$ - So, money lasts about 27 full years before depletion. **Final answers:** - (a) $A_1 = 1.004 x$, $A_2 = 1.004^2 x + 1.004 x$ - (b)(i) $A_3 = 1.004^3 x + 1.004^2 x + 1.004 x$, $A_4 = 1.004^4 x + 1.004^3 x + 1.004^2 x + 1.004 x$ - (b)(ii) $A_{120} = 251 (1.004^{120} - 1) x$ - (c) $A_{216} = 251 (1.004^{216} - 1) x$ - (d) Minimum $x = 59$ - (e) Money lasts approximately 28 years after starting withdrawals.