一位著名的艺术家想要创作一件只使用正方形的艺术品。第一天，他画了一个边长为 $x$ 米的正方形的周长，如图 1 所示。第二天，他将该正方形分成 9 个相等的正方形，并画出中间正方形的周长，如图 2 所示。第三天，他将第二天剩下的 8 个空正方形分别分成 9 个相等的正方形，并画出这些中间正方形的周长，如图 3 所示。在随后的日子里重复这种“画出中间正方形周长”的步骤（图 4 展示了他第四天完成的作品）。

1. Write down the number of new squares drawn on Day 3 and Day 4.

On Day 2, there are $8$ empty squares remaining.
On Day 3, each of these $8$ squares is divided, and $1$ middle square is drawn in each.
Number of new squares on Day 3 $= 8$

This leaves $8 \times 8 = 64$ empty squares.
On Day 4, each of the $64$ empty squares is divided, and $1$ middle square is drawn in each.
Number of new squares on Day 4 $= 64$

2. Formulate an expression for the number of new squares on Day $N$.

For $N \ge 2$, the number of new squares drawn each day forms the sequence $1, 8, 64, \dots$
We can express these as consecutive powers of $8$:
Day 2 ($N=2$): $1 = 8^0$
Day 3 ($N=3$): $8 = 8^1$
Day 4 ($N=4$): $64 = 8^2$

Following this pattern, the exponent is always $N - 2$. Thus, for Day $N$, the number of new squares is $8^{N-2}$.

1. Find the side lengths for Day 3 and Day 4.

Each day, the squares from the previous day are divided into a 3 by 3 grid, so the side length of a new square is $\frac{1}{3}$ of the side length of a square from the previous day.
Length on Day 2 = $\frac{x}{3}$ m.
Length on Day 3 = $\frac{1}{3} \times \frac{x}{3} = \frac{x}{9}$ m.
Length on Day 4 = $\frac{1}{3} \times \frac{x}{9} = \frac{x}{27}$ m.

2. Generalise the side length to find the formula for Day $N$.

The side lengths follow a geometric progression:
Day 2 ($N=2$): $\frac{x}{3^1}$
Day 3 ($N=3$): $\frac{x}{3^2}$
Day 4 ($N=4$): $\frac{x}{3^3}$
Therefore, on Day $N$ (for $N \ge 2$), the power of $3$ in the denominator is $N-1$. The side length is $\frac{x}{3^{N-1}}$ m.

1. State the expressions for the number of squares and the side length of each square on Day $N$ from earlier parts.

From (i), the number of new squares on Day $N$ is $8^{N-2}$.
From (ii), the side length of a new square on Day $N$ is $\frac{x}{3^{N-1}}$ m.

2. Formulate the expression for the total perimeter of the new squares.

$\text{Total perimeter} = (\text{Number of squares}) \times 4 \times (\text{side length of one square})$
$= 8^{N-2} \times 4 \times \frac{x}{3^{N-1}}$

3. Use the laws of indices to rearrange the expression into the required form.

$= \frac{8^{N-1}}{8^1} \times 4 \times \frac{x}{3^{N-1}}$
$= \frac{4}{8} \times x \times \frac{8^{N-1}}{3^{N-1}}$
$= \frac{1}{2}x \left(\frac{8}{3}\right)^{N-1}$
$= \frac{x}{2}\left(\frac{8}{3}\right)^{N-1} \quad \text{(Shown)}$

1. Identify that the total perimeter requires adding the Day 1 perimeter to the sum of perimeters from Day 2 to Day $N$.

\text{Perimeter drawn on Day 1} = 4x

\text{Total perimeter} = 4x + \sum_{k=2}^{N} \frac{x}{2}\left(\frac{8}{3}\right)^{k-1}

2. Calculate the sum of the geometric progression for the perimeters from Day 2 to Day $N$.

\text{The perimeters from Day 2 to } N ext{ form a geometric progression with:}

a = \frac{x}{2}\left(\frac{8}{3}\right)^{2-1} = \frac{4x}{3}

r = \frac{8}{3}

\text{Number of terms } n = N - 1

S_{n} = \frac{a(r^{n} - 1)}{r - 1} = \frac{\frac{4x}{3} \left( \left(\frac{8}{3}\right)^{N-1} - 1 \right)}{\frac{8}{3} - 1}

S_{n} = \frac{\frac{4x}{3} \left( \left(\frac{8}{3}\right)^{N-1} - 1 \right)}{\frac{5}{3}} = \frac{4x}{5} \left( \left(\frac{8}{3}\right)^{N-1} - 1 \right)

3. Combine the Day 1 perimeter with the GP sum and simplify.

\text{Total perimeter} = 4x + \frac{4x}{5} \left( \left(\frac{8}{3}\right)^{N-1} - 1 \right)

= \frac{20x}{5} - \frac{4x}{5} + \frac{4x}{5}\left(\frac{8}{3}\right)^{N-1}

= \frac{16x}{5} + \frac{4x}{5}\left(\frac{8}{3}\right)^{N-1}

= \frac{4x}{5} \left( 4 + \left(\frac{8}{3}\right)^{N-1} \right)

1. Find the total number of squares drawn by the artist by the end of Day 6.

Based on the pattern, the artist draws $1$ square on Day 1. For Day $N$ ($N \ge 2$), the artist draws $8^{N-2}$ new squares.
By the end of Day 6, the total number of squares drawn is:
$1 + (8^0 + 8^1 + 8^2 + 8^3 + 8^4)$
Using the sum formula for a geometric progression:
$1 + \frac{1(8^5 - 1)}{8 - 1} = 1 + \frac{32768 - 1}{7}$
$= 1 + 4681 = 4682 \text{ squares}$

2. Determine the number of squares the director personally pays for.

The director pays for the $1^\text{st}, 4^\text{th}, 7^\text{th}, \dots$ squares. Their positions form an arithmetic progression with a general term of $m_k = 3k - 2$.
To find the maximum number of squares $k$ the director pays for:
$3k - 2 \le 4682$
$3k \le 4684 \implies k \le 1561.33\dots$
Since $k$ must be an integer, $k = 1561$.
The director pays for $1561$ squares (the last being the $4681^\text{st}$ square).

3. Formulate the sequence of payments made by the director.

The pledge for the $m$-th square is $10 + (m - 1)(1) = m + 9$ dollars.
The amounts paid by the director form an arithmetic progression:
First payment ($m=1$): $\char36{}10$
Second payment ($m=4$): $\char36{}13$
Third payment ($m=7$): $\char36{}16$
This sequence is an arithmetic progression with first term $a = 10$ and common difference $d = 3$.

4. Calculate the total amount paid by the director.

We use the sum of an arithmetic progression formula $S_n = \frac{n}{2}[2a + (n-1)d]$ with $n = 1561$, $a = 10$, and $d = 3$:
$\text{Total amount} = \frac{1561}{2} \left[ 2(10) + (1561 - 1)(3) \right]$
$= \frac{1561}{2} \left[ 20 + 4680 \right]$
$= \frac{1561}{2} (4700)$
$= 1561 \times 2350 = 3668350$
The director will personally pay $\char36{}3,668,350$.