Jane decides to purchase an apartment. To finance her apartment, she intends to apply for a bank housing loan of $847 500 which charges a monthly interest rate of 0.25%. Interest is charged on the outstanding loan on the first day of each month, starting from the second month. A fixed monthly repayment of $M is done on the last day of each month throughout the loan tenure, starting from the first month.

1. Determine the loan balance at the end of the first month.

Let the loan balance at the end of the $n$th month be $B_n$.
For the first month, no interest is charged. A repayment of $\char36{}M$ is made at the end of the month:
$B_1 = 847500 - M$

2. Determine the loan balance at the end of the second month.

For the second month, interest of $0.25\%$ is charged on $B_1$ at the start of the month, and another repayment of $\char36{}M$ is made at the end:
$B_2 = 1.0025B_1 - M$
$B_2 = 1.0025(847500 - M) - M$
$B_2 = 847500(1.0025) - M(1.0025) - M$
$B_2 = 847500(1.0025) - M(1 + 1.0025)$
$B_2 = 849618.75 - 2.0025M$

3. Establish the pattern for the loan balance at the end of the $n$th month.

Continuing this pattern for the third month:
$B_3 = 1.0025B_2 - M = 847500(1.0025)^2 - M(1 + 1.0025 + 1.0025^2)$
Generalising to the $n$th month:
$B_n = 847500(1.0025)^{n-1} - M(1 + 1.0025 + 1.0025^2 + \dots + 1.0025^{n-1})$

4. Evaluate the sum of the geometric series formed by the repayment terms.

The terms multiplying $M$ form a geometric series with first term $a = 1$, common ratio $r = 1.0025$, and $n$ terms.
$\text{Sum} = \frac{a(r^n - 1)}{r - 1}$
$\text{Sum} = \frac{1(1.0025^n - 1)}{1.0025 - 1}$
$\text{Sum} = \frac{1.0025^n - 1}{0.0025}$
$\text{Sum} = 400(1.0025^n - 1)$

5. Substitute the sum back into the expression to show the final result.

Substitute this sum back into the expression for $B_n$:
$B_n = 847500(1.0025)^{n-1} - M[400(1.0025^n - 1)]$
$B_n = 847500(1.0025)^{n-1} - 400M(1.0025^n - 1)$
$B_n = 847500(1.0025)^{n-1} + 400M(1 - 1.0025^n) \quad \text{(shown)}$

1. Substitute the total number of months into the loan balance formula and equate it to zero.

The loan tenure is 25 years, which equates to $n = 25 \times 12 = 300$ months.
At the end of the 300th month, the loan is fully repaid, so the loan balance is $0$. Substituting $n = 300$ into the expression from part (a):
$847500(1.0025)^{299} + 400M\left(1 - 1.0025^{300}\right) = 0$

2. Solve for the fixed monthly repayment $M$.

$400M\left(1.0025^{300} - 1\right) = 847500(1.0025)^{299}$
$M = \frac{847500(1.0025)^{299}}{400\left(1.0025^{300} - 1\right)}$
$M \approx 4011.944$

3. Calculate the TDSR and conclude whether the bank will approve the loan.

Assuming $M$ is Jane's only monthly debt, her total monthly debt is $\char36{}4011.944$.
$\text{TDSR} = \frac{4011.944}{7300} \times 100 \approx 54.958\%$
Since her TDSR of $54.96\%$ is less than or equal to $55\%$, the bank will approve Jane's loan.

1. State a valid assumption made in calculating the TDSR for the loan approval

In part (b), the calculated housing loan monthly repayment $M$ was substituted directly as the 'Borrower's total monthly debt'. For this to be accurate, we assume that Jane has no other existing financial debts (e.g., car loans, credit card debts).

(Alternatively, it is also acceptable to state that the bank's monthly interest rate of 0.25% is assumed to remain constant throughout the entire 25-year tenure.)