A famous artist wants to create an artwork that uses squares only. On the first day, he draws the perimeter of a square with length $x$ m as shown in Figure 1. On the second day, he divides the square into 9 equal squares and draws the perimeter of the middle square as shown in Figure 2. On the third day, he divides the remaining 8 empty squares from Day 2 into 9 equal squares, and draws the perimeters of the middle squares as shown in Figure 3. This procedure of "drawing the perimeters of the middle squares" is then repeated for subsequent days (see Figure 4 for his finished effort on Day 4).

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 $\char36{}3,668,350$.