Chapter 6.5, Problem 42E

To determine

To write a sequence that represents the number of teams that have been eliminated after $n$ rounds of the badminton tournament.

Answer to Problem 42E

The sequences of geometric progression that have been eliminated after $n$ rounds of the badminton tournament is, ${\text{a}}_n=96\cdot {\left(4\right)}^{n-1}$

The sequences are $96,384,1536,6144,32768,\mathrm{......}$

Explanation of Solution

Given: The graphing calculator screen displays an area 96 square units. After you zoom out once, the area is 384 square units. After you zoom out second time, the area is 1536 square units.

The given sequences are $96,384,1536,\mathrm{..........}$

The given sequences are in geometric progression.

For $nth$ term the geometric progression is,

  ${\text{a}}_n=a_1{r}^{n-1}$

First term, ${\text{a}}_1=96$

Common ratio, $\text{r}=\frac{384}{96}=4$

Therefore,

The sequences of geometric progression that have been eliminated after $n$ rounds of the badminton tournament is ${\text{a}}_n=96\cdot {\left(4\right)}^{n-1}$

Now,

Second term, ${\text{a}}_2=a_{1}r$

  $\begin{array}{l}{\text{a}}_2=(96)\cdot (4)\\ {\text{a}}_2=384\end{array}$

Third term,

  $\begin{array}{l}{\text{a}}_3=a_2{r}^2\\ {\text{a}}_3=96\cdot {\left(4\right)}^2\\ {\text{a}}_3=96\cdot \left(16\right)\\ {\text{a}}_3=1536\end{array}$

Fourth term,

  $\begin{array}{l}{\text{a}}_4=a_1{r}^3\\ {\text{a}}_4=96\cdot {\left(4\right)}^3\\ {\text{a}}_4=96\cdot \left(64\right)\\ {\text{a}}_4=6144\end{array}$

Fifth term,

  $\begin{array}{l}{\text{a}}_5=a_1{r}^4\\ {\text{a}}_5=128\cdot {\left(4\right)}^4\\ {\text{a}}_5=128\cdot \left(256\right)\\ {\text{a}}_5=32768\end{array}$

Hence,

The sequences of geometric progression that have been eliminated after $n$ rounds of the badminton tournament is, ${\text{a}}_n=96\cdot {\left(4\right)}^{n-1}$

The sequences are $96,384,1536,6144,32768,\mathrm{......}$