Chapter 7.3, Problem 60PS

Solution

i.

To calculation: The rule for $a_n$ then find the total number of squares removed through stage 8.

The Sierpinski carpet is a fractal created using squares. The process involves removing smaller squares from larger squares. First. divide a large square into nine congruent squares. Remove the center square. Repeat these steps for each smaller square, as shown below. Assume that each side of the initial square is one unit long.

  

The required rule is $a_n=8^{n-1}$ and sum of $s_8=2396745$ .

Given information:

The given terms are $a_1=1\text{and}r=8$ .

Calculation:

The number of squares removed at the first stage $a_1=1$

The number of squares removed at the second stage $a_2=8$

  $\begin{array}{l}r=\frac{a_2}{a_1}\\ r=\frac{8}{1}\end{array}$

It is given that $a_1=1\text{and}r=8$ .

The general formula for a geometric sequence is $a_n=a_1\cdot r^{n-1}$ .

Put the value of $a_1=1\text{and}r=8$ into $a_n=a_1\cdot r^{n-1}$

  $\begin{array}{l}{a}_n=1\cdot 8^{n-1}\\ a_n=8^{n-1}\end{array}$

Therefore, the required rule $a_n=8^{n-1}$

The total number of squares removed through stage $8$

Put the value of $n=8\text{and}r=8$ into $s_n=a_1\frac{1-r^n}{1-r}$

  $\begin{array}{l}{s}_n=a_1\frac{1-r^n}{1-r}\\ s_8=1\cdot \frac{1-8^8}{1-8}\\ s_8=\frac{1-8^8}{7}\\ s_8=2396745\end{array}$ .

ii.

To identify: The rule for b. Then find the remaining area of the original square after stage 12. Let b, be the remaining area of the original square after the nth stage.

The rule of $b_n$ is $b_n={\left(\frac{8}{9}\right)}^{n}s$ and the value of 12 stage is $b_n=0.27s$ .

Explanation:

Consider the given sequence.

After the first stage the remaining area is

  $\begin{array}{l}{b}_1=s-\frac{1}{9}s\\ b_1=s\left(1-\frac{1}{9}\right)\\ b_1=\frac{8}{9}s\end{array}$

After the second stage the remaining area is

  $\begin{array}{l}{b}_2=\frac{8}{9}s-8\cdot \frac{1}{9^2}s\\ b_2=\left(\frac{8}{9}s-\frac{8}{9^2}\right)s\\ b_2=\left(\frac{8\cdot 9}{9^2}-\frac{8}{9^2}\right)s\\ b_2=\left(\frac{8\cdot \left(9-1\right)}{9^2}\right)\\ b_2={\left(\frac{8}{9}\right)}^{2}s\end{array}$

Observing this, it can conclude that the rule is $b_n={\left(\frac{8}{9}\right)}^{n}s$

  $\begin{array}{l}{b}_n={\left(\frac{8}{9}\right)}^{12}s\\ b_n=0.27s\end{array}$