The diagrams below show a sequence of patterns formed by squares. Stage 0 is represented by a square of length 1. At each successive stage, squares with half the length of the smallest square in the previous stage are added to the unoccupied vertices of the squares in the previous stage. Let $A_0$ be the area of the square in Stage 0 and $A_n$ be the area of each new square added in Stage $n$. For example, $A_0 = 1$ and $A_1 = \frac{1}{4}$.

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:
$\text{Total squares} = 1 + 13120 = 13121$