下图展示了由正方形组成的一系列图案。阶段 0 由一个边长为 1 的正方形表示。在随后的每个阶段中，将前一阶段中最小正方形边长一半的正方形，添加到前一阶段中正方形的未占用顶点上。 设 $A_0$ 为阶段 0 的正方形面积， $A_n$ 为阶段 $n$ 新添加的每个正方形的面积。例如， $A_0 = 1$ 且 $A_1 = \frac{1}{4}$。

1. Determine the side length of the squares added in Stage $n$.

Let the side length of the squares added in Stage $n$ be $L_n$.
Since the side length is halved at each successive stage and $L_0 = 1$, we have a geometric sequence:
$L_n = \left(\frac{1}{2}\right)^n$

2. Calculate the area $A_n$ in terms of $n$.

The area of each square is the square of its side length:
$A_n = (L_n)^2$
$A_n = \left[ \left(\frac{1}{2}\right)^n \right]^2$
$A_n = \left(\frac{1}{2}\right)^{2n} = \left(\frac{1}{4}\right)^n$

1. Determine the area of each new square added at Stage $n$ (from part a).

Since the side length of the square halves at each stage, the area of the square becomes $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ of the previous stage's square.
With $A_0 = 1$ and $A_1 = \frac{1}{4}$, the area of each new square at Stage $n$ is:
$A_n = \left(\frac{1}{4}\right)^n$

2. Determine the number of new squares added at Stage $n$.

At Stage 1, $4$ new squares are added.
In each subsequent stage, new squares are added to the $3$ unoccupied vertices of each square added in the previous stage.
Thus, the number of new squares multiplies by $3$ at each stage, forming a geometric sequence with first term $4$ and common ratio $3$:
Number of new squares at Stage $n = 4 \times 3^{n-1}$

3. Calculate the total area of the new squares added at Stage $n$.

Total area = (Number of new squares) $\times A_n$
$= (4 \times 3^{n-1}) \times \left(\frac{1}{4}\right)^n$
$= 4 \times 3^{n-1} \times \frac{1}{4 \times 4^{n-1}}$
$= \frac{3^{n-1}}{4^{n-1}}$
$= \left(\frac{3}{4}\right)^{n-1} \text{ (Shown)}$

1. Express the total area of Stage $n$ as the sum of the initial Stage 0 area and the areas added in all subsequent stages up to $n$.

$\text{Total Area} = 1 + \sum_{k=1}^n \left(\frac{3}{4}\right)^{k-1}$

2. Calculate the sum of the new areas added from Stage 1 to Stage $n$ using the geometric series sum formula.

The series $1 + \frac{3}{4} + \left(\frac{3}{4}\right)^2 + \dots + \left(\frac{3}{4}\right)^{n-1}$ is a geometric progression with first term $a = 1$, common ratio $r = \frac{3}{4}$, and $n$ terms.

$S_n = \frac{1\left[1 - \left(\frac{3}{4}\right)^n\right]}{1 - \frac{3}{4}}$
$S_n = \frac{1 - \left(\frac{3}{4}\right)^n}{\frac{1}{4}} = 4\left[1 - \left(\frac{3}{4}\right)^n\right]$

3. Add the area of Stage 0 and simplify to obtain the final expression.

$\text{Total Area} = 1 + 4\left[1 - \left(\frac{3}{4}\right)^n\right]$
$= 1 + 4 - 4\left(\frac{3}{4}\right)^n$
$= 5 - 4\left(\frac{3}{4}\right)^n \text{ (Shown)}$

1. Give a mathematical reason for convergence

As $n \to \infty$, the term $\left(\frac{3}{4}\right)^n$ approaches $0$ because the common ratio $r = \frac{3}{4}$ satisfies $-1 < r < 1$.

2. Calculate the value of the converging area

Using the expression from part (c):
$\text{Total area} = \lim_{n \to \infty} \left[ 5 - 4\left(\frac{3}{4}\right)^n \right]$
$= 5 - 4(0) = 5$

1. Set up the inequality for the area at Stage $m$ using the area of the square fractal.

From part (d), the area of the square fractal is $5$.
The total area in Stage $m$ must exceed 90% of this value:
$5 - 4\left(\frac{3}{4}\right)^m > 0.9 \times 5$
$5 - 4(0.75)^m > 4.5$

2. Solve the inequality to find the integer value of $m$.

$4(0.75)^m < 0.5$
$0.75^m < 0.125$
Taking natural logarithms on both sides:
$m \ln(0.75) < \ln(0.125)$
Since $\ln(0.75)$ is negative, the inequality sign flips:
$m > \frac{\ln(0.125)}{\ln(0.75)}$
$m > 7.228$
Since $m$ must be an integer, $m = 8$.

3. Identify the pattern for the number of squares added at each stage.

The number of new squares added at each stage forms a sequence:
Stage 1: $4$ squares
Stage 2: $12$ squares
Stage 3: $36$ squares
This is a geometric progression with first term $a = 4$ and common ratio $r = 3$.

4. Calculate the total number of squares at Stage 8.

The sum of new squares added from Stage 1 to Stage 8 is:
$S_8 = \frac{4(3^8 - 1)}{3 - 1} = 2(6561 - 1) = 13120$
The total number of squares includes the initial $1$ square at Stage 0:
$\text{Total squares} = 1 + 13120 = 13121$