Chapter 8.3, Problem 61E

a.

To determine

To find: The rule for total number of square removed at $n\text{th}$ stage $a_n$ .

Explanation of Solution

Given information:

Below diagram represents first three iteration of Sierpinski carpet.

Calculation:

Sirpinski carpet is created by first dividing each square in $9$ parts and then removing center square. Again repeat the steps for each remaining squares.

From above diagram, first stage $1$ square was removed, at second stage $8$ square were removed, third stage $64$ squares were removed and so on. i.e. $a_1=1,r=8$ .

  $\begin{array}{l}\text{So, }{a}_n=1{\left(8\right)}^{n-1}\text{ where }{a}_n=\text{Number of square removed at }n\text{th stage}\text{.}\\ a_1=1\\ a_2=8\\ a_3=64\\ a_4=8^{4-1}=8^3=512\\ a_5=8^{5-1}=8^4=4096\\ a_6=8^{6-1}=8^5=32768\\ a_7=8^{7-1}=8^6=262144\\ a_8=8^{8-1}=8^7=2097152\end{array}$

Hence, $a_n=1{\left(8\right)}^{n-1}\text{ where }{a}_n=\text{Number of square removed at }n\text{th stage}$ .

b.

To determine

To find: The rule for remaining area $b_n$ of original area after $n\text{th}$ stage .

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

Sirpinski carpet is created by first dividing each square in $9$ parts and then removing center square. Again repeat the steps for each remaining squares.

When we divide the base square into $9$ smaller equal sized square of area $\frac{1}{9}$ and remove center square. The remaining area will be $\frac{8}{9}$ .

In next stage $2$ , each smaller square of area $\frac{1}{9}$ is divided into $9$ equal parts and the area of each part will be $\frac{1}{9}\cdot \frac{1}{9}$ . There are total $8\cdot 8$ smaller square with each area $\frac{1}{9}\cdot \frac{1}{9}$ remaining. Therefore, area of remaining part will be $8\cdot 8\cdot \frac{1}{9}\cdot \frac{1}{9}={\left(\frac{8}{9}\right)}^2$ .

Further stages follow same pattern.

Hence, area of remaining part will be given as −

  $b_n={\left(\frac{8}{9}\right)}^n,\text{ where }{b}_n\text{ is the area of remaining part at }n\text{th stage}\text{.}$

At stage $12$ ,

  $b_{12}={\left(\frac{8}{9}\right)}^{12}=0.2433$ .