Predicting What Comes Next: Sequences and Progressions | Exercise 8.3

Question 7

Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet. Stage 0 of this fractal is a square sheet of paper. To construct Stage 1, each side of the square is trisected and the points of trisection of opposite sides are joined to obtain nine smaller squares. The centre square is then removed and the 8 smaller squares are retained, leaving a square hole in the centre. The same process is repeated on the eight smaller shaded squares to obtain Stage 2 and so on.

Look at Fig. 8.12 and try to answer the following questions.

(i) How many red squares are there in Stages 0 to 3? (ii) Can you predict the number of red squares in Stages 4 and 5? (iii) Can you find a rule for the number of red squares at the $n^{\text{th}}$ stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage. (iv) Suppose the area of the square in Stage 0 is 1 square unit. What is the area of the red region in Stages 1, 2 and 3? What will be the area of the red region in Stages 4 and 5? Find the explicit as well as the recursive formula for the area of the red region at the $n^{\text{th}}$ stage. What happens to this area as $n$ , the number of stages, goes on increasing?

Solution

Understand the Question

The Sierpiński carpet is constructed recursively starting from a unit square (Stage 0).
At each step, every retained square is divided into $9$ equal smaller squares and the central square is removed, leaving $8$ squares behind.
Thus, at each subsequent stage:
- The number of red squares is multiplied by $8$ .
- The area of the red region is multiplied by $\dfrac{8}{9}$ .

(i) How many red squares are there in Stages 0 to 3?

Step 1 · Count Red Squares in Stages 0 to 3

Diagram 1

At Stage 0: $\text{Number of red squares} = 1$

At Stage 1 (divided into $9$ squares, $1$ center square removed): $\text{Number of red squares} = 8$

At Stage 2 (each of the $8$ squares yields $8$ smaller squares):

\begin{aligned} \text{Number of red squares} &= 8 \times 8 \\ &= 8^2 = 64 \end{aligned}

At Stage 3:

\begin{aligned} \text{Number of red squares} &= 64 \times 8 \\ &= 8^3 = 512 \end{aligned}

Answer

(i) Stage 0: $1$ , Stage 1: $8$ , Stage 2: $64$ , Stage 3: $512$

(ii) Can you predict the number of red squares in Stages 4 and 5?

Step 1 · Calculate Number of Squares for Stages 4 and 5

At Stage 4:

\begin{aligned} \text{Number of red squares} &= 8^4 \\ &= 512 \times 8 \\ &= 4096 \end{aligned}

At Stage 5:

\begin{aligned} \text{Number of red squares} &= 8^5 \\ &= 4096 \times 8 \\ &= 32768 \end{aligned}

Answer

(ii) Stage 4: $4096$ , Stage 5: $32768$

(iii) Can you find a rule for the number of red squares at the $n^{\text{th}}$ stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage.

Step 1 · Formulate Explicit and Recursive Rules

Let $t_n$ be the number of red squares at stage $n$ for $n \ge 0$ .

Explicit formula: $t_n = 8^n$

Recursive formula: $t_0 = 1 \quad \text{and} \quad t_n = 8 \times t_{n-1} \text{ for } n \ge 1$

Answer

(iii) Explicit formula: $t_n = 8^n$ ; Recursive formula: $t_0 = 1, t_n = 8t_{n-1} \text{ for } n \ge 1$

(iv) Suppose the area of the square in Stage 0 is 1 square unit. What is the area of the red region in Stages 1, 2 and 3? What will be the area of the red region in Stages 4 and 5? Find the explicit as well as the recursive formula for the area of the red region at the $n^{\text{th}}$ stage. What happens to this area as $n$ , the number of stages, goes on increasing?

Step 1 · Find Area for Stages 1, 2, and 3

Given the area at Stage 0 is $A_0 = 1$ . Since $\dfrac{1}{9}$ of the area is removed at each step, $\dfrac{8}{9}$ of the previous area remains.

Stage 1: $\text{Area} = 1 \times \dfrac{8}{9} = \dfrac{8}{9}$

Stage 2: $\text{Area} = \left(\dfrac{8}{9}\right)^2 = \dfrac{64}{81}$

Stage 3: $\text{Area} = \left(\dfrac{8}{9}\right)^3 = \dfrac{512}{729}$

Step 2 · Find Area for Stages 4 and 5

Stage 4: $\text{Area} = \left(\dfrac{8}{9}\right)^4 = \dfrac{4096}{6561}$

Stage 5: $\text{Area} = \left(\dfrac{8}{9}\right)^5 = \dfrac{32768}{59049}$

Step 3 · Determine Formulas and Long-Term Behavior

Explicit formula: $A_n = \left(\dfrac{8}{9}\right)^n$

Recursive formula: $A_0 = 1 \quad \text{and} \quad A_n = \dfrac{8}{9} \times A_{n-1} \text{ for } n \ge 1$

Behavior as $n \to \infty$ : Since the common ratio $\dfrac{8}{9} < 1$ , as $n$ increases, the area of the red region approaches $0$ .

Answer

(iv) Areas: Stage 1 = $\dfrac{8}{9}$ , Stage 2 = $\dfrac{64}{81}$ , Stage 3 = $\dfrac{512}{729}$ , Stage 4 = $\dfrac{4096}{6561}$ , Stage 5 = $\dfrac{32768}{59049}$ . Explicit formula: $A_n = \left(\dfrac{8}{9}\right)^n$ . Recursive formula: $A_0 = 1, A_n = \dfrac{8}{9}A_{n-1}$ for $n \ge 1$ . As $n \to \infty$ , the area approaches $0$ .

Common Mistakes

Base Case Indexing: Note that the sequence begins at Stage $0$ ( $n=0$ ), meaning $t_0 = 8^0 = 1$ and $A_0 = \left(\frac{8}{9}\right)^0 = 1$ . Writing $t_n = 8^{n-1}$ would be an off-by-one indexing error.
Confusing Count and Area Ratios: The number of squares grows by a factor of $8$ at each step, but the total area is multiplied by $\dfrac{8}{9}$ because each new square has $\dfrac{1}{9}$ the area of the previous square.

Question 7

More questions in Exercise 8.3