在某家银行，一个储蓄账户在每个月的最后一天按每月 0.9% 的利率支付利息。2024年1月1日，Mr Monie 在储蓄账户中存入了 $10000 的初始存款，并在之后的每个月开始时继续存入 $810。

1. Calculate the total amount at the end of the first month after interest is earned.

$10000 \times \left(1 + \frac{0.9}{100}\right) = 10000 \times 1.009 = 10090$

2. Add the deposit for the second month and calculate the amount at the end of the second month after interest.

$(10090 + 810) \times 1.009 = 10900 \times 1.009 = 10998.1$

1. Establish a pattern for the total amount at the end of the $n^{\text{th}}$ month.

Let $A_n$ be the total amount at the end of the $n^{\text{th}}$ month.
At the end of the 1st month:
$A_1 = 10000(1.009)$

At the end of the 2nd month:
$A_2 = [10000(1.009) + 810](1.009) = 10000(1.009)^2 + 810(1.009)$

At the end of the 3rd month:
$A_3 = [A_2 + 810](1.009) = 10000(1.009)^3 + 810(1.009)^2 + 810(1.009)$

By observing the pattern, the total amount at the end of the $n^{\text{th}}$ month is:
$A_n = 10000(1.009)^n + 810(1.009)^{n-1} + 810(1.009)^{n-2} + \dots + 810(1.009)$

2. Identify the geometric series and compute its sum.

The terms involving subsequent deposits form a geometric series:
$810(1.009) + 810(1.009)^2 + \dots + 810(1.009)^{n-1}$
This series has $n - 1$ terms, with first term $a = 810(1.009) = 817.29$ and common ratio $r = 1.009$.

Using the sum of a finite geometric series, $S = \frac{a(r^{n-1} - 1)}{r - 1}$:
$S = \frac{817.29(1.009^{n-1} - 1)}{1.009 - 1}$
$S = \frac{817.29}{0.009}(1.009^{n-1} - 1)$
$S = 90810(1.009^{n-1} - 1)$

3. Substitute the sum back and simplify to the required form.

Substitute the sum back into the expression for $A_n$:
$A_n = 10000(1.009)^n + 90810(1.009^{n-1} - 1)$

Rewrite $10000(1.009)^n$ as $10000(1.009)(1.009^{n-1})$ to factor out $1.009^{n-1}$:
$A_n = 10090(1.009^{n-1}) + 90810(1.009^{n-1}) - 90810$
$A_n = (10090 + 90810)(1.009^{n-1}) - 90810$
$A_n = 100900(1.009^{n-1}) - 90810$

$\therefore$ The amount at the end of the $n^{\text{th}}$ month is $\char36{}100900(1.009)^{n-1} - \char36{}90810$ (shown).

1. Set up and solve an inequality to find when the end-of-month balance exceeds \$30000.

$100900(1.009)^{n-1} - 90810 > 30000$
$100900(1.009)^{n-1} > 120810$
$(1.009)^{n-1} > \frac{12081}{10090}$
$n - 1 > \frac{\ln(12081/10090)}{\ln 1.009} \approx 20.10$
$n > 21.10$

2. Evaluate the balance at the end of the 21st month and at the start of the 22nd month to find the precise crossing point.

$\text{At the end of the 21st month } (n=21)\text{:}$
$\text{Balance} = 100900(1.009)^{20} - 90810 \approx \$29903.94$
$\text{This is less than } \$30000 \text{, so the threshold is not crossed by the end of month 21.}$
$\text{At the start of the 22nd month, a deposit of } \$810 \text{ is made:}$
$\text{Balance} = 29903.94 + 810 = \$30713.94$
$\text{Since } \$30713.94 > \$30000\text{, the balance first exceeds } \$30000 \text{ at the start of the 22nd month.}$

3. Determine the calendar date for the start of the 22nd month.

$\text{Month 1 is January 2024.}$
$\text{Month 22 is } 21 \text{ months after January 2024.}$
$21 \text{ months} = 1 \text{ year and } 9 \text{ months.}$
$\text{January } 2024 + 1 \text{ year} = \text{January } 2025.$
$\text{January } 2025 + 9 \text{ months} = \text{October } 2025.$
$\text{Since deposits are made at the start of each month, the date is 1 October 2025.}$

1. Identify the disbursements as an arithmetic progression and formulate the sum for $N$ months.

The disbursements form an arithmetic progression with first term $a = 100$ and common difference $d = 10$.
Sum of disbursements, $S_N = \frac{N}{2}[2(100) + (N-1)(10)]$
$S_N = \frac{N}{2}[200 + 10N - 10]$
$S_N = \frac{N}{2}[10N + 190]$
$S_N = 5N^2 + 95N$

2. Add the sum of disbursements to the initial principal to find the total amount.

$\text{Total amount} = 30000 + S_N$
$\text{Total amount} = 30000 + 5N^2 + 95N$

1. Formulate the expression for the total amount in the savings account after $N$ months.

Let $A_1$ be the total amount in the savings account after $N$ months. Since there are no further deposits, the initial amount of $\char36{}30000$ simply accrues compound interest at $0.9\%$ per month.
$A_1 = 30000(1.009)^N$

2. Set up the inequality for the financial product to be more profitable.

Let $A_2$ be the total amount in the financial product after $N$ months. From part (d), the initial $\char36{}30000$ is added to the sum of an arithmetic progression of disbursements ($a=100, d=10$).
$A_2 = 30000 + \frac{N}{2}[2(100) + (N-1)(10)]$
$A_2 = 30000 + 5N^2 + 95N$
For the financial product to be more profitable, we require $A_2 > A_1$:
$30000 + 5N^2 + 95N > 30000(1.009)^N$

3. Solve the inequality using a graphing calculator.

Using a graphing calculator, we find the points of intersection of the graphs $Y_1 = 30000 + 5N^2 + 95N$ and $Y_2 = 30000(1.009)^N$.
The intersections occur at $N \approx 48.38$ and $N \approx 343.45$.
From the graphs, $Y_1 > Y_2$ when $48.38 < N < 343.45$.

4. Deduce the integer range for $N$.

Since $N$ represents the number of months, it must take on integer values.
The range of values of $N$ is therefore:
$49 \le N \le 343$