A first order recurrence relation for a sequence $u_0, u_1, u_2, \dots$ is an equation that relates the general term $u_n$ with its preceding term $u_{n-1}$. A student takes up a university study loan of $20\,000$ from a bank. The bank charges a $0.5\%$ interest rate per month. The interest is added on at the first day of each month. The student decides to make repayments monthly on the last day of each month. Each repayment is $450 except for the final repayment which will ultimately pay off the loan. Denoting the amount that he owes the bank at the end of the $n$th month by $u_n$, the above scenario can be represented by the following first order recurrence relation, $u_n = 1.005\,u_{n-1} - 450, \quad n \ge 1.$

1. Calculate $u_1$ by substituting $n=1$ and the initial value $u_0$ into the recurrence relation.

$u_1 = 1.005\,u_0 - 450$
$u_1 = 1.005(20\,000) - 450$
$u_1 = 20\,100 - 450 = 19\,650$

2. Calculate $u_2$ using the value of $u_1$.

$u_2 = 1.005\,u_1 - 450$
$u_2 = 1.005(19\,650) - 450$
$u_2 = 19\,748.25 - 450 = 19\,298.25$

3. Calculate $u_3$ using the value of $u_2$.

$u_3 = 1.005\,u_2 - 450$
$u_3 = 1.005(19\,298.25) - 450$
$u_3 = 19\,394.74125 - 450 = 18\,944.74125$
As this represents a monetary amount, rounding to $2$ decimal places gives $18\,944.74$, though the exact value is acceptable.

1. Iterate the recurrence relation to establish the emerging algebraic pattern.

$u_n = 1.005 u_{n-1} - 450$

Substituting $u_{n-1} = 1.005 u_{n-2} - 450$:
$u_n = 1.005(1.005 u_{n-2} - 450) - 450$
$u_n = 1.005^2 u_{n-2} - 450(1.005) - 450$

Substituting $u_{n-2} = 1.005 u_{n-3} - 450$:
$u_n = 1.005^2(1.005 u_{n-3} - 450) - 450(1.005) - 450$
$u_n = 1.005^3 u_{n-3} - 450(1.005^2) - 450(1.005) - 450$
$u_n = 1.005^3 u_{n-3} - 450(1.005^2 + 1.005 + 1)$

2. Generalise the pattern to the $n$-th step, expressing $u_n$ in terms of $u_0$.

Repeating this process $n$ times yields:
$u_n = 1.005^n u_0 - 450(1.005^{n-1} + 1.005^{n-2} + \dots + 1.005 + 1)$

3. Use the sum of a geometric progression to simplify the series portion.

The series $(1 + 1.005 + \dots + 1.005^{n-1})$ is a geometric progression with first term $a = 1$, common ratio $r = 1.005$, and $n$ terms.

$S_n = \frac{a(r^n - 1)}{r - 1} = \frac{1(1.005^n - 1)}{1.005 - 1} = \frac{1.005^n - 1}{0.005}$

4. Substitute the value of $u_0$ and simplify to obtain the final required form.

Given $u_0 = 20000$:
$u_n = 20000(1.005^n) - 450 \left( \frac{1.005^n - 1}{0.005} \right)$
$u_n = 20000(1.005^n) - 90000(1.005^n - 1)$
$u_n = 20000(1.005^n) - 90000(1.005^n) + 90000$
$u_n = 90000 - 70000(1.005^n) \quad \text{(shown)}$

1. Set up an inequality to determine when the loan amount drops to zero or below.

$u_n \le 0$
$90000 - 1.005^n(70000) \le 0$
$1.005^n \ge \frac{90000}{70000}$
$1.005^n \ge \frac{9}{7}$

2. Solve for $n$ to find the number of months required.

$n \ln(1.005) \ge \ln\left(\frac{9}{7}\right)$
$n \ge \frac{\ln(9/7)}{\ln(1.005)}$
$n \ge 50.388$
Since payments are made monthly, 51 months are required to fully repay the loan (as $u_{50} > 0$).

3. Calculate the outstanding loan balance at the end of the 50th month.

After 50 full regular payments of $\char36{}450$, the outstanding amount is:
$u_{50} = 90000 - 1.005^{50}(70000)$
$u_{50} \approx 90000 - 70000(1.283226)$
$u_{50} \approx 174.1637$

4. Calculate the final repayment amount in the 51st month.

In the 51st month, the interest is added to the remaining balance before the final clearing repayment is made.
$\text{Final repayment} = 1.005 \times u_{50}$
$= 1.005 \times 174.1637$
$= 175.0345$
$\approx \$175.03$