Chapter 8.3, Problem 59E

a.

To determine

To find: The rule for the number of games played in nth round of the given sequence.

Answer to Problem 59E

  $a_n=32{\left(\frac{1}{2}\right)}^{n-1}$ ; $n\le 6$ .

Explanation of Solution

Given information:

In the regional soccer tournament $32$ games were played in first round and in subsequent rounds number of games decreased by a factor of $\frac{1}{2}$ .

Calculation:

In the regional soccer tournament $32$ games were played in first round and in subsequent rounds number of games decreased by a factor of $\frac{1}{2}$ .

i.e. $a_1=32,r=\frac{1}{2}$ .

General rule for nth term of a geometric series is given by-

  $a_n=a_1{r}^{n-1}$

Substitute $n=1,r=\frac{1}{2}$ ,

Therefore the rule for nth term of given geometric series would be given by −

  $a_n=32{\left(\frac{1}{2}\right)}^{n-1}$ .

  $\begin{array}{l}\text{For }n=6,a_6=32{\left(\frac{1}{2}\right)}^{6-1}=1\\ \text{For }n=7,a_7=32{\left(\frac{1}{2}\right)}^{7-1}=\frac{1}{2}\end{array}$

Since number of games cannot be a fraction $\left(<1\right)$ . Thus total $6$ rounds of games were played in the tournament. This rule for number of games in nth round makes sense for $n\le 6$ .

b.

To determine

To find: The total number of games played in the regional tournament.

Answer to Problem 59E

  $63$ .

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

From part $a$ of this exercise-

  $a_1=32,r=\frac{1}{2}$ , $a_n=32{\left(\frac{1}{2}\right)}^{n-1}$ and $n\le 6$ .

Using sum formula for geometric progression,

  $\begin{array}{l}{S}_n=a_1\left(\frac{1-r^n}{1-r}\right)\\ \text{Substitute }n=6,r=\frac{1}{2},a_1=32\\ S_6=a_1\left(\frac{1-{\left(\frac{1}{2}\right)}^6}{1-\frac{1}{2}}\right)\\ \Rightarrow S_6=32\times \frac{63}{32}\\ \Rightarrow S_6=63\end{array}$

Hence, total number of games played in the tournament were $63$ .