QUESTION 7
Easy
The given 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 Stage 0 to 3?
(ii) Can you predict the number of red squares in Stage 4 and 5?
(iii) Can you find the rule for the number of red squares at the \(n^{th}\) 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 1? 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 \(n^{th}\) stage. What happens to this area as \(n\), the number of stages, goes on increasing?
SOLUTION
1
(i) How many red squares are there in stage 0 to 3?
At Stage 0, there is only one square.
\(R_0 = 1\)
At Stage 1, the center square is removed, leaving 8 squares.
\(R_1 = 8\)
At Stage 2, each of the 8 red squares produces 8 new red squares. So
\(R_2 = 8 × 8 = 64\)
At Stage 3, each of the 64 red squares produces 8 new red squares.
\(R_3 = 64 × 8 = 512\)
\(R_0 = 1\)
At Stage 1, the center square is removed, leaving 8 squares.
\(R_1 = 8\)
At Stage 2, each of the 8 red squares produces 8 new red squares. So
\(R_2 = 8 × 8 = 64\)
At Stage 3, each of the 64 red squares produces 8 new red squares.
\(R_3 = 64 × 8 = 512\)
2
(ii) Predict the number of red squares in Stage 4 and 5
Since each stage multiplies by 8,
In Stage 4, \(R_4 = 512 × 8 = 4096\)
There are 4096 red squares.
In Stage 5, \(R_5 = 4096 × 8 = 32768\)
There are 32768 red squares.
In Stage 4, \(R_4 = 512 × 8 = 4096\)
There are 4096 red squares.
In Stage 5, \(R_5 = 4096 × 8 = 32768\)
There are 32768 red squares.
3
(iii) Find a rule for the number of red squares at the \(n^{th}\) stage
The sequence is: 1, 8, 64, 512, ...
This is a GP where
\(a = 1,\, r = 8\)
For a GP, \(T_n = ar^{n−1}\)
Since stages begin at Stage 0, the explicit formula is
\(R_n = 8^n\)
Recursive formula:
\(R_0 = 1\)
\(R_n = 8R_{n−1}\), \(n≥1\)
This is a GP where
\(a = 1,\, r = 8\)
For a GP, \(T_n = ar^{n−1}\)
Since stages begin at Stage 0, the explicit formula is
\(R_n = 8^n\)
Recursive formula:
\(R_0 = 1\)
\(R_n = 8R_{n−1}\), \(n≥1\)
4
(iv) Area of the red region
At Stage 0, \(A_0 = 1\)
At Stage 1, the square is divided into 9 equal parts. One part(center) is removed, So the remaining area is
\(A_1 = \frac{8}{9}\)
At Stage 2, again, only \(\frac{8}{9}\) of the previous area remains.
\(A_2 = \frac{8}{9} × \frac{8}{9} = (\frac{8}{9})^2 = \frac{64}{81}\)
At Stage 3,
\(A_3 = (\frac{8}{9})^3 = \frac{512}{729}\)
At Stage 1, the square is divided into 9 equal parts. One part(center) is removed, So the remaining area is
\(A_1 = \frac{8}{9}\)
At Stage 2, again, only \(\frac{8}{9}\) of the previous area remains.
\(A_2 = \frac{8}{9} × \frac{8}{9} = (\frac{8}{9})^2 = \frac{64}{81}\)
At Stage 3,
\(A_3 = (\frac{8}{9})^3 = \frac{512}{729}\)
5
Find area in Stages 4 and 5
At Stage 4, \(A_4 = (\frac{8}{9})^4 = \frac{4096}{6561}\)
At Stage 5, \(A_5 = (\frac{8}{9})^5 = \frac{32768}{59049}\)
At Stage 5, \(A_5 = (\frac{8}{9})^5 = \frac{32768}{59049}\)
6
Write explicit and recursive formula for area
Th areas form a GP: \(1,\, (\frac{8}{9}),\,(\frac{8}{9})^2,\,(\frac{8}{9})^3,...\)
with \(a = 1,\, r = \frac{8}{9}\)
Hence, the explicit formula is
\(A_n = (\frac{8}{9})^n\)
And recursive formula:
\(A_0 = 1\)
\(A_n = \frac{8}{9} A_{n−1}\), for \(n≥1\)
with \(a = 1,\, r = \frac{8}{9}\)
Hence, the explicit formula is
\(A_n = (\frac{8}{9})^n\)
And recursive formula:
\(A_0 = 1\)
\(A_n = \frac{8}{9} A_{n−1}\), for \(n≥1\)
7
What happens as \(n\) increases?
Since \(0 < \frac{8}{9} < 1\), repeated multiplication by \(\frac{8}{9}\) makes the area smaller and smaller.
Concept Note
A geometric progression(GP) is a sequence where each term is obtained by multiplying the previous term by a fixed number called the common ratio(r).
For any GP with \(0 < r < 1\), the terms get closer and closer to zero.
Thus the Sierpiński carpet keeps developing more holes, and the remaining red area approaches 0.