The figure 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 the figure 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 nth 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 nth stage. What happens to this area as n, the number of stages, goes on increasing?
Step-by-step solution
Idea: At each stage every red square is split into 9 equal squares and the centre one is removed. So the number of red squares is multiplied by 8 and the red area is multiplied by 89: both are GPs.
(i) How many red squares are there in Stages 0 to 3?
- Stage 0: 1. Each red square becomes 8 red squares at the next stage.½ mark
- Stage 1: 8; Stage 2: 8 × 8 = 64; Stage 3: 64 × 8 = 512.½ mark
(ii) Can you predict the number of red squares in Stages 4 and 5?
- Stage 4: 512 × 8 = 4096; Stage 5: 4096 × 8 = 32768.1 mark
(iii) Can you find a rule for the number of red squares at the nth stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage.
- 1 = 80, 8 = 81, 64 = 82, 512 = 83: explicit formula tn = 8n.½ mark
- Recursive formula: t0 = 1, tn = 8 × tn−1 for n ≥ 1 (the stages are numbered from 0).½ mark
- (If the stages are counted from 1, with t1 = 1 for Stage 0, the same rule is written t1 = 1, tn = 8tn−1 for n ≥ 2; then tn = 8n−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 nth stage. What happens to this area as n, the number of stages, goes on increasing?
- Each stage keeps 8 of the 9 equal small squares of every red square, so the red area is multiplied by 89 at every stage.½ mark
- Stage 1: 89; Stage 2: 6481; Stage 3: 512729; Stage 4: 40966561; Stage 5: 3276859049 square units.½ mark
- Explicit: sn = (89)n. Recursive: s0 = 1, sn = 89 × sn−1 for n ≥ 1.½ mark
- Since 89 < 1, the area keeps decreasing: about 0.89, 0.79, 0.70, 0.62, 0.55, … It decreases slowly but gets closer and closer to 0 as n increases (it never becomes exactly 0).½ mark
Check: Stage 2 area: 64 small squares, each (19)2 = 181 of the original: 64 × 181 = 6481 ✓.
Answer to write in the exam
(i)
Each red square → 8 red squares at the next stage
∴ Stages 0 to 3: 1, 8, 64, 512
(ii)
Stage 4 = 512 × 8 = 4096
Stage 5 = 4096 × 8 = 32768
(iii)
1 = 80, 8 = 81, 64 = 82, 512 = 83
Explicit: tn = 8n
Recursive: t0 = 1, tn = 8tn−1 for n ≥ 1
(iv)
Area at each stage = 89 × area at previous stage
Stages 1, 2, 3: 89, 6481, 512729 sq. units
Stages 4, 5: 40966561, 3276859049 sq. units
Explicit: sn = (89)n
Recursive: s0 = 1, sn = 89sn−1 for n ≥ 1
89 < 1 ∴ the area decreases towards 0 as n increases.
Common mistakes that cost marks
- Counting 9 squares at Stage 1. The centre square is removed, so 8 remain.
- Writing the area at Stage 2 as 89 − 19 = 79. The area is multiplied by 89 again: 6481.
- Mixing the two numberings: writing tn = 8n together with t1 = 1. If Stage 0 is t0, start the rule at t0 = 1.
How this can come in the exam
In the Sierpiński carpet (Stage 0 area 1), the red area at Stage 2 is
- 6481
- 1618
- 79
- 881
Show answer
(A) 6481
(89)2 = 6481.
A Sierpiński carpet is made from a red square of side 27 cm. Find the side of each red square and the total red area at Stage 3.
Show answer
Each stage divides the side by 3: 27 ÷ 33 = 1 cm (1 mark). Number of squares 83 = 512, so red area = 512 cm2 (1 mark).Try one yourself
An equilateral triangle is divided into 9 equal smaller triangles and 3 of them are removed; this is repeated on every remaining triangle. Starting from area 1, write the explicit formula for the number of triangles and the area at Stage n.
Show answer
6 triangles are kept each time: number tn = 6n; area sn = (69)n = (23)n.
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