在 2023 年 1 月 1 日，金先生因其业务需要向银行贷款 $\char36{}40000$。银行为还款提供了两种方案。在方案 A 下，银行收取 $\char36{}4660$ 的一次性管理费，该费用计入贷款总额中。金先生需要在每个月月初支付 $\char36{}700$ 的月供，且每往后一个月，月供金额增加 $\char36{}60$，直到贷款全部还清。

1. Identify the parameters of the arithmetic progression representing the sequence of payments.

The monthly instalments form an arithmetic progression with the first term $a = 700$ and common difference $d = 60$.

2. Apply the sum formula for an arithmetic progression and simplify to obtain the given expression.

$\begin{aligned}S_k &= \frac{k}{2} [2a + (k-1)d] \\ &= \frac{k}{2} [2(700) + (k-1)(60)] \\ &= \frac{k}{2} [1400 + 60k - 60] \\ &= \frac{k}{2} [60k + 1340] \\ &= 30k^2 + 670k \end{aligned}$

1. Set up the equation by equating the expression for the total amount paid to the total loan amount.

Total loan amount = $\char36{}40000 + \char36{}4660 = \char36{}44660$

$30k^2 + 670k = 44660$

2. Solve the quadratic equation to find the value of $k$.

Divide by 10:
$3k^2 + 67k - 4466 = 0$

Factorise the quadratic expression:
$(3k + 223)(k - 29) = 0$

$k = -\frac{223}{3}$ (reject since $k > 0$) or $k = 29$

It will take $29$ payments to fully repay the loan.

1. Calculate the outstanding amount after the first month (end of January 2023).

Amount after Jan = $40000(1.015) - p = 40600 - p$

2. Calculate the outstanding amount after the second month (end of February 2023), which is the amount owed on 1st Mar 2023.

Amount on 1st Mar = $(40600 - p)(1.015) - p$
$= 41209 - 1.015p - p$
$= 41209 - 2.015p$

1. Establish the pattern for the amount owed at the end of the first few months.

Let $A_n$ be the amount owed at the end of the $n^{\text{th}}$ month.
Given the interest rate of $1.5\%$, the multiplier is $1 + 0.015 = 1.015$.
$A_1 = 40000(1.015) - p$
$A_2 = [40000(1.015) - p](1.015) - p = 40000(1.015)^2 - 1.015p - p$
$A_3 = [40000(1.015)^2 - 1.015p - p](1.015) - p = 40000(1.015)^3 - 1.015^2p - 1.015p - p$

2. Generalise the expression for the $n^{\text{th}}$ month and identify the geometric progression.

By observing the pattern, the amount owed at the end of the $n^{\text{th}}$ month is:
$A_n = 40000(1.015)^n - p(1.015^{n-1} + 1.015^{n-2} + \dots + 1.015 + 1)$

The series in the brackets is a geometric progression with first term $a = 1$, common ratio $r = 1.015$, and number of terms $= n$.
Using the sum of geometric series formula $S_n = \frac{a(r^n - 1)}{r - 1}$:
Sum $= \frac{1(1.015^n - 1)}{1.015 - 1} = \frac{1.015^n - 1}{0.015} = \frac{200}{3}(1.015^n - 1)$

3. Substitute the geometric sum back into the expression and find $\alpha$ and $\beta$.

Substitute the sum back into $A_n$:
$A_n = 40000(1.015)^n - \frac{200}{3}p(1.015^n - 1)$

Comparing this with the given expression $40000\alpha^{n}-\beta p(\alpha^{n}-1)$:
$\alpha = 1.015$
$\beta = \frac{200}{3}$

1. Evaluate the formula for the outstanding amount $A_n$ after $n$ standard payments, using the parameters derived from the loan conditions.

From (ii)(b), the amount owed after $n$ months is $A_n = 40000\alpha^n - \beta p(\alpha^n-1)$.
Given the interest rate of $1.5\%$ per month, $\alpha = 1.015$. The sum of the geometric progression gives $\beta = \frac{1}{0.015} = \frac{200}{3}$.
Substituting $p = 1585$:
$A_n = 40000(1.015)^n - \frac{200}{3}(1585)(1.015^n - 1)$
$A_n = 40000(1.015)^n - \frac{317000}{3}(1.015^n - 1)$

2. Relate the final payment $m$ to the remaining balance $A_k$ and set up an inequality.

At the end of the $k^{\text{th}}$ month, the amount owed before paying $m$ is $1.015 A_{k-1}$, which is exactly $m$.
If a standard payment of $\char36{}1585$ were made in the $k^{\text{th}}$ month, the remaining balance would be $A_k = 1.015 A_{k-1} - 1585 = m - 1585$.
Since Mr Kim pays off the loan with a final payment $0 < m \le 1585$, we have:
$A_k \le 0$
$40000(1.015)^k - \frac{317000}{3}(1.015^k - 1) \le 0$

3. Solve the inequality to find the number of months $k$.

$\left(40000 - \frac{317000}{3}\right)(1.015)^k + \frac{317000}{3} \le 0$
$-\frac{197000}{3}(1.015)^k \le -\frac{317000}{3}$
$(1.015)^k \ge \frac{317}{197}$
Taking the natural logarithm on both sides:
$k \ln(1.015) \ge \ln\left(\frac{317}{197}\right)$
$k \ge \frac{\ln(317/197)}{\ln(1.015)} \approx 31.949$
Since $k$ must be an integer, $k = 32$.

4. Calculate the exact outstanding amount $m$ paid at the end of the $32^{\text{nd}}$ month.

When $k = 32$, the balance after $31$ standard payments is:
$A_{31} = 40000(1.015)^{31} - \frac{317000}{3}(1.015^{31} - 1) \approx 1434.70448$
The outstanding amount $m$ is this balance plus the $1.5\%$ interest charged in the $32^{\text{nd}}$ month:
$m = 1.015 \times A_{31} \approx 1.015(1434.70448) \approx 1456.225$
Rounding to the nearest cent, $m = 1456.23$.