1. **Problem statement:** Mary's grandparents invest an amount $x$ on 1st January 1998 and continue to deposit $x$ on the first day of each month. The account pays 0.4% interest per month, compounded monthly on the last day of each month. We define $A_n$ as the amount in the account on the last day of the $n$th month after interest is added. 2. **Part (a): Find $A_1$ and show $A_2 = 1.004^2 x + 1.004 x$** - At the end of the first month, the initial deposit $x$ earns interest 0.4%, so $$A_1 = x \times 1.004 = 1.004 x.$$ - On the first day of the second month, another $x$ is deposited, so before interest at the end of month 2, the amount is $$A_1 + x = 1.004 x + x = (1.004 + 1) x = 2.004 x.$$ - Interest is then added at 0.4%, so $$A_2 = 2.004 x \times 1.004 = 1.004^2 x + 1.004 x,$$ which matches the required expression. 3. **Part (b)(i): Expressions for $A_3$ and $A_4$** - At the start of month 3, deposit $x$ again, so before interest, $$A_2 + x = (1.004^2 x + 1.004 x) + x = 1.004^2 x + 1.004 x + x.$$ - After interest at month 3, $$A_3 = (1.004^2 x + 1.004 x + x) \times 1.004 = 1.004^3 x + 1.004^2 x + 1.004 x.$$ - Similarly, at the start of month 4, deposit $x$ again, so before interest, $$A_3 + x = 1.004^3 x + 1.004^2 x + 1.004 x + x.$$ - After interest at month 4, $$A_4 = (1.004^3 x + 1.004^2 x + 1.004 x + x) \times 1.004 = 1.004^4 x + 1.004^3 x + 1.004^2 x + 1.004 x.$$ 4. **Part (b)(ii): Show amount before 10th birthday is $251 (1.004^{120} - 1) x$** - The amount after $n$ months is a geometric series: $$A_n = x(1.004^n + 1.004^{n-1} + \cdots + 1.004) = x \sum_{k=1}^n 1.004^k.$$ - Using the formula for the sum of a geometric series, $$\sum_{k=1}^n r^k = r \frac{r^n - 1}{r - 1},$$ with $r=1.004$, we get $$A_n = x \times 1.004 \frac{1.004^n - 1}{1.004 - 1} = x \times 1.004 \frac{1.004^n - 1}{0.004}.$$ - Simplify the denominator, $$\frac{1.004}{0.004} = 251,$$ so $$A_n = 251 x (1.004^n - 1).$$ - Mary turns 10 years old after 120 months (10 years × 12 months), so $$A_{120} = 251 x (1.004^{120} - 1),$$ which is the required expression. **Final answers:** - $A_1 = 1.004 x$ - $A_2 = 1.004^2 x + 1.004 x$ - $A_3 = 1.004^3 x + 1.004^2 x + 1.004 x$ - $A_4 = 1.004^4 x + 1.004^3 x + 1.004^2 x + 1.004 x$ - Amount before 10th birthday: $251 (1.004^{120} - 1) x$