2016年5月初，萨姆（Sam）向银行借了 $50 000，银行在每个月末收取 0.2% 的特惠月利率。萨姆从 2016 年 6 月开始，在每个月初偿还每期 $1 000 的分期付款。

1. Formulate the expression for the amount owed at the end of the month after the 1st instalment is paid (June 2016).

Amount owed at the end of May = $50\ 000(1.002)$

Let $A_1$ be the amount owed at the end of June.
$A_1 = [50\ 000(1.002) - 1\ 000](1.002)$
$A_1 = 50\ 000(1.002)^2 - 1\ 000(1.002)$

2. Find the expression for the amount owed at the end of the next month to establish a pattern.

Let $A_2$ be the amount owed at the end of July.
$A_2 = (A_1 - 1\ 000)(1.002)$
$A_2 = [50\ 000(1.002)^2 - 1\ 000(1.002) - 1\ 000](1.002)$
$A_2 = 50\ 000(1.002)^3 - 1\ 000(1.002)^2 - 1\ 000(1.002)$

3. Generalise the pattern for the amount owed at the end of the month after the $n$th instalment.

Let $A_n$ be the amount owed at the end of the month after the $n$th instalment.
$A_n = 50\ 000(1.002)^{n+1} - 1\ 000(1.002)^n - \dots - 1\ 000(1.002)^2 - 1\ 000(1.002)$
$A_n = 50\ 000(1.002)^{n+1} - 1\ 000[1.002^n + 1.002^{n-1} + \dots + 1.002^2 + 1.002]$

4. Apply the geometric progression sum formula to simplify the expression and obtain the required result.

The series inside the brackets is a geometric progression with $n$ terms, first term $a = 1.002$, and common ratio $r = 1.002$.
$S_n = \frac{a(r^n - 1)}{r - 1} = \frac{1.002(1.002^n - 1)}{1.002 - 1} = \frac{1.002(1.002^n - 1)}{0.002} = 501(1.002^n - 1)$

Substitute the sum back into $A_n$:
$A_n = 50\ 000(1.002)^{n+1} - 1\ 000 \times 501(1.002^n - 1)$
$A_n = 50\ 000(1.002^{n+1}) - 501\ 000(1.002^n - 1)\ \text{(shown)}$

1. Set up an inequality to find when the amount owed becomes zero or less.

$50\ 000(1.002^{n+1}) - 501\ 000(1.002^n - 1) \le 0$

2. Simplify the inequality to isolate $1.002^n$.

$50\ 000(1.002)(1.002^n) - 501\ 000(1.002^n) + 501\ 000 \le 0$
$50\ 100(1.002^n) - 501\ 000(1.002^n) \le -501\ 000$
$-450\ 900(1.002^n) \le -501\ 000$
$1.002^n \ge \frac{501\ 000}{450\ 900}$
$1.002^n \ge \frac{10}{9}$

3. Solve for $n$ using logarithms and deduce the integer number of instalments.

$n \ln(1.002) \ge \ln\left(\frac{10}{9}\right)$
$n \ge \frac{\ln(10/9)}{\ln 1.002}$
$n \ge 52.73$
Since $n$ must be a whole number, 53 instalments are required.

1. Find the total number of instalments needed by setting the amount owed to less than or equal to 0 (from part ii).

From part (ii), we require the number of instalments $n$ such that the remaining amount is $\le 0$:
$50000(1.002^{n+1}) - 501000(1.002^n - 1) \le 0$
$50100(1.002^n) - 501000(1.002^n) + 501000 \le 0$
$-450900(1.002^n) \le -501000$
$1.002^n \ge \frac{5010}{4509} \implies n \ge 52.8$
Thus, 53 instalments are required to settle the amount.

2. Formulate the expression for the amount paid on the last (53rd) instalment.

The 53rd instalment will exactly clear the outstanding balance at the end of the 52nd month. We substitute $n = 52$ into the given formula for the amount owed to find the outstanding balance right before the final instalment is paid:
$\text{Amount to pay} = U_{52}$
$= 50000(1.002^{53}) - 501000(1.002^{52} - 1)$

3. Evaluate the remaining balance to the nearest cent.

$= 55576.4826 - 54764.8256$
$= 811.657$
$\approx 811.66$
The last instalment is $811.66.

1. Adapt the expression from part (i) to form an inequality for a new monthly instalment of $x$ over 19 months.

Let $x$ be the minimum amount paid each month.
From part (i), for an instalment of $\char36{}1000$, the amount owed at the end of the month after the $n$th instalment is $50\ 000(1.002^{n+1}) - 501\ 000(1.002^n - 1)$.
The constant $501\ 000$ comes from the sum of the geometric progression of instalments: $1000 \times \frac{1.002}{0.002} = 1000 \times 501$.
Thus, for an instalment of $x$, the amount owed at the end of the month after the 19th instalment ($n = 19$) is:
$50\ 000(1.002^{20}) - 501x(1.002^{19} - 1)$

2. Solve the inequality to find the minimum whole dollar amount.

To settle the amount owed, the remaining balance must be less than or equal to $0$:
$50\ 000(1.002^{20}) - 501x(1.002^{19} - 1) \le 0$
$501x(1.002^{19} - 1) \ge 50\ 000(1.002^{20})$
$x \ge \frac{50\ 000(1.002^{20})}{501(1.002^{19} - 1)}$
$x \ge 2685.559\dots$
Rounding to the nearest dollar, the minimum amount Sam should pay is $\char36{}2686$.