On 1st Jan 2023, Mr Kim borrows $\char36{}40000$ for his business from a bank. The bank provides two options for repaying the loan. Under Plan A, the bank imposes a one-time administrative fee of $\char36{}4660$, which is added to the loan amount. Mr Kim needs to pay a monthly installment of $\char36{}700$ at the start of each month, and the installment amount increases by $\char36{}60$ for each subsequent month, until the loan amount is fully paid.

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, $m = 1456.23$.