Chapter 6.5, Problem 41E

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 41E

The sequences of geometric progression that have been eliminated after $n$ rounds of the badminton tournament, ${\text{a}}_n=128\cdot {\left(\frac{1}{2}\right)}^{n-1}$

The sequences are $128,64,32,16,8,4,\mathrm{....}$

Explanation of Solution

Given: A badminton tournament begins with 128 teams. After the first round, 64 teams remain. After the second round, 32 team remain, so on.

The given sequences are $128,64,32,\mathrm{......}$

The given sequences are in geometric progression.

For $nth$ term the geometric progression is,

  ${\text{a}}_n=a_1{r}^{n-1}$

First term, ${\text{a}}_1=128$

Common ratio, $\text{r}=\frac{64}{128}=\frac{1}{2}$

Therefore,

The sequences of geometric progression that have been eliminated after $n$ rounds of the badminton tournament is,

  ${\text{a}}_n=128\cdot {\left(\frac{1}{2}\right)}^{n-1}$

Now,

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

  $\begin{array}{l}{\text{a}}_2=128\cdot \left(\frac{1}{2}\right)\\ {\text{a}}_2=64\end{array}$

Third term,

  $\begin{array}{l}{\text{a}}_3=a_2{r}^2\\ {\text{a}}_3=128\cdot {\left(\frac{1}{2}\right)}^2\\ {\text{a}}_3=128\cdot {\left(\frac{1}{2}\right)}^2\\ {\text{a}}_3=128\cdot \left(\frac{1}{4}\right)\\ {\text{a}}_3=32\end{array}$

Fourth term,

  $\begin{array}{l}{\text{a}}_4=a_1{r}^3\\ {\text{a}}_4=128\cdot {\left(\frac{1}{2}\right)}^3\\ {\text{a}}_4=128\cdot \left(\frac{1}{8}\right)\\ {\text{a}}_4=16\end{array}$

Fifth term,

  $\begin{array}{l}{\text{a}}_5=a_1{r}^4\\ {\text{a}}_5=128\cdot {\left(\frac{1}{2}\right)}^4\\ {\text{a}}_5=128\cdot \left(\frac{1}{16}\right)\\ {\text{a}}_5=8\end{array}$

Sixth term,

  $\begin{array}{l}{\text{a}}_6=a_1{r}^5\\ {\text{a}}_6=128\cdot {\left(\frac{1}{2}\right)}^5\\ {\text{a}}_6=128\cdot \left(\frac{1}{32}\right)\\ {\text{a}}_6=4\end{array}$

Hence,

The sequences of geometric progression that have been eliminated after $n$ rounds of the badminton tournament is, ${\text{a}}_n=128\cdot {\left(\frac{1}{2}\right)}^{n-1}$

The sequence will be $128,64,32,16,8,4,\mathrm{....}$