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,
$u_n = 1.005u_{n-1} - 450, \quad n \ge 1.$

1. Use the recurrence relation and $u_0$ to find $u_1$.

$u_1 = 1.005u_0 - 450$
$u_1 = 1.005(20000) - 450 = 20100 - 450 = 19650$

2. Use the recurrence relation and $u_1$ to find $u_2$.

$u_2 = 1.005u_1 - 450$
$u_2 = 1.005(19650) - 450 = 19748.25 - 450 = 19298.25$

3. Use the recurrence relation and $u_2$ to find $u_3$, giving the answer to an appropriate degree of accuracy for money.

$u_3 = 1.005u_2 - 450$
$u_3 = 1.005(19298.25) - 450 = 19394.74125 - 450 = 18944.74125$
$\approx 18944.74$ (to 2 d.p.)

1. Express the general term $u_n$ in terms of $n$ using the sum of a geometric progression.

By expanding the recurrence relation:
$u_1 = 1.005(20000) - 450$
$u_2 = 1.005^2(20000) - 450(1.005) - 450$
Following the pattern,
$u_n = 20000(1.005)^n - 450(1 + 1.005 + \dots + 1.005^{n-1})$
Using the formula for the sum of a finite geometric series ($S_n = a\frac{r^n - 1}{r - 1}$):
$u_n = 20000(1.005)^n - 450 \left( \frac{1.005^n - 1}{1.005 - 1} \right)$
$u_n = 20000(1.005)^n - 90000(1.005^n - 1)$
$u_n = 90000 - 70000(1.005)^n$

2. Set up and solve an inequality to find the number of months required to repay the loan.

To repay the loan, the amount owed must reach zero or fall below zero:
$u_n \le 0$
$90000 - 70000(1.005)^n \le 0$
$70000(1.005)^n \ge 90000$
$1.005^n \ge \frac{9}{7}$
Taking the natural logarithm on both sides:
$n \ln 1.005 \ge \ln \left(\frac{9}{7}\right)$
$n \ge \frac{\ln(9/7)}{\ln 1.005} \approx 50.39$
Since $n$ represents whole months, $n = 51$ months.

3. Calculate the final repayment amount.

At the end of the 50th month, the remaining loan balance is:
$u_{50} = 90000 - 70000(1.005)^{50} \approx 174.156$
For the 51st month, interest is added before the repayment. Thus, the final repayment required to exactly clear this balance is:
$\text{Final repayment} = 1.005 \times u_{50} = 1.005 \times 174.156 \approx 175.03$

(Alternatively, if a full $\char36{}450$ payment was hypothetically made, $u_{51} = 90000 - 70000(1.005)^{51} \approx -274.97$, indicating an overpayment of $\char36{}274.97$. The true final repayment is therefore $450 - 274.97 = \char36{}175.03$.)