Chapter 9, Problem 7PS

(a)

To determine

To write: The first 20 terms of the given sequence.

Answer to Problem 7PS

The first 20 terms are 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1.

Explanation of Solution

Given:

The piecewise-defined sequence is,

  $a_1=7$ , $a_n=\{\begin{array}{c}\frac{a_{n-1}}{2},\text{ when }{a}_{n-1}\text{ is even}\text{.}\\ 3a_{n-1}+1,\text{ when }{a}_{n-1}\text{ is odd}\text{.}\end{array}$

Calculation:

To determine the first 20 terms of the sequence, put $n=2,\dots ,20$ as shown below.

  $\begin{array}{c}{a}_2=3a_1+1\left(\because a_1=7\text{ is odd}\right)\\ =3\left(7\right)+1\\ =22\end{array}$

  $\begin{array}{c}{a}_3=\frac{a_2}{2}\left(\because a_2=22\text{ is even}\right)\\ =\frac{22}{2}\\ =11\end{array}$

  $\begin{array}{c}{a}_4=3a_3+1\left(\because a_3=11\text{ is odd}\right)\\ =3\left(11\right)+1\\ =34\end{array}$

  $\begin{array}{c}{a}_5=\frac{a_4}{2}\left(\because a_4=34\text{ is even}\right)\\ =\frac{34}{2}\\ =17\end{array}$

  $\begin{array}{c}{a}_6=3a_5+1\left(\because a_5=17\text{ is odd}\right)\\ =3\left(17\right)+1\\ =52\end{array}$

  $\begin{array}{c}{a}_7=\frac{a_6}{2}\left(\because a_6=52\text{ is even}\right)\\ =\frac{52}{2}\\ =26\end{array}$

  $\begin{array}{c}{a}_8=\frac{a_7}{2}\left(\because a_7=26\text{ is even}\right)\\ =\frac{26}{2}\\ =13\end{array}$

  $\begin{array}{c}{a}_9=3a_8+1\left(\because a_5=13\text{ is odd}\right)\\ =3\left(13\right)+1\\ =40\end{array}$

  $\begin{array}{c}{a}_{10}=\frac{a_9}{2}\left(\because a_9=40\text{ is even}\right)\\ =\frac{40}{2}\\ =20\end{array}$

  $\begin{array}{c}{a}_{11}=\frac{a_{10}}{2}\left(\because a_{10}=20\text{ is even}\right)\\ =\frac{20}{2}\\ =10\end{array}$

  $\begin{array}{c}{a}_{12}=\frac{a_{11}}{2}\left(\because a_{11}=10\text{ is even}\right)\\ =\frac{10}{2}\\ =5\end{array}$

  $\begin{array}{c}{a}_{13}=3a_{12}+1\left(\because a_{12}=5\text{ is odd}\right)\\ =3\left(5\right)+1\\ =16\end{array}$

  $\begin{array}{c}{a}_{14}=\frac{a_{13}}{2}\left(\because a_{13}=16\text{ is even}\right)\\ =\frac{16}{2}\\ =8\end{array}$

  $\begin{array}{c}{a}_{15}=\frac{a_{14}}{2}\left(\because a_{14}=8\text{ is even}\right)\\ =\frac{8}{2}\\ =4\end{array}$

  $\begin{array}{c}{a}_{16}=\frac{a_{15}}{2}\left(\because a_{15}=4\text{ is even}\right)\\ =\frac{4}{2}\\ =2\end{array}$

  $\begin{array}{c}{a}_{17}=\frac{a_{16}}{2}\left(\because a_{16}=2\text{ is even}\right)\\ =\frac{2}{2}\\ =1\end{array}$

  $\begin{array}{c}{a}_{18}=3a_{17}+1\left(\because a_{17}=1\text{ is odd}\right)\\ =3\left(1\right)+1\\ =4\end{array}$

  $\begin{array}{c}{a}_{19}=\frac{a_{18}}{2}\left(\because a_{18}=4\text{ is even}\right)\\ =\frac{4}{2}\\ =2\end{array}$

  $\begin{array}{c}{a}_{20}=\frac{a_{19}}{2}\left(\because a_{19}=2\text{ is even}\right)\\ =\frac{2}{2}\\ =1\end{array}$

So, the first 20 terms are 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1.

(b)

To determine

To write: The first 10 terms of the sequences for which $a_1=4$ , $a_1=5$ and $a_1=12$ .

Answer to Problem 7PS

The first 10 terms of the sequence for which $a_1=4$ are 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, for $a_1=5$ the terms are 5, 16, 8, 4, 2, 1, 4, 2, 1, 4 and for $a_1=12$ are 12, 6, 3, 10, 5, 16, 8, 4, 2, 1. Each sequence ends in 4, 2, 1 on repeat.

Explanation of Solution

Calculation:

To determine the first 10 terms of the sequence for which $a_1=4$ , put $n=2,\dots ,10$ as shown below:

  $\begin{array}{c}{a}_2=\frac{a_1}{2}\left(\because a_1=4\text{ is even}\right)\\ =\frac{4}{2}\\ =2\end{array}$

  $\begin{array}{c}{a}_3=\frac{a_2}{2}\left(\because a_2=2\text{ is even}\right)\\ =\frac{2}{2}\\ =1\end{array}$

  $\begin{array}{c}{a}_4=3a_3+1\left(\because a_3=1\text{ is odd}\right)\\ =3\left(1\right)+1\\ =4\end{array}$

  $\begin{array}{c}{a}_5=\frac{a_4}{2}\left(\because a_4=4\text{ is even}\right)\\ =\frac{4}{2}\\ =2\end{array}$

  $\begin{array}{c}{a}_6=\frac{a_5}{2}\left(\because a_5=2\text{ is even}\right)\\ =\frac{2}{2}\\ =1\end{array}$

  $\begin{array}{c}{a}_7=3a_6+1\left(\because a_6=1\text{ is odd}\right)\\ =3\left(1\right)+1\\ =4\end{array}$

  $\begin{array}{c}{a}_8=\frac{a_7}{2}\left(\because a_7=4\text{ is even}\right)\\ =\frac{4}{2}\\ =2\end{array}$

  $\begin{array}{c}{a}_9=\frac{a_8}{2}\left(\because a_8=2\text{ is even}\right)\\ =\frac{2}{2}\\ =1\end{array}$

  $\begin{array}{c}{a}_{10}=3a_9+1\left(\because a_9=1\text{ is odd}\right)\\ =3\left(1\right)+1\\ =4\end{array}$

The obtained sequence is 4, 2, 1, 4, 2, 1, 4, 2, 1, 4.

To determine the first 10 terms of the sequence for which $a_1=5$ , put $n=2,\dots ,10$ as shown below:

  $\begin{array}{c}{a}_2=3a_1+1\left(\because a_1=5\text{ is odd}\right)\\ =3\left(5\right)+1\\ =16\end{array}$

  $\begin{array}{c}{a}_3=\frac{a_2}{2}\left(\because a_2=16\text{ is even}\right)\\ =\frac{16}{2}\\ =8\end{array}$

  $\begin{array}{c}{a}_4=\frac{a_3}{2}\left(\because a_3=8\text{ is even}\right)\\ =\frac{8}{2}\\ =4\end{array}$

  $\begin{array}{c}{a}_5=\frac{a_4}{2}\left(\because a_4=4\text{ is even}\right)\\ =\frac{4}{2}\\ =2\end{array}$

  $\begin{array}{c}{a}_6=\frac{a_5}{2}\left(\because a_5=2\text{ is even}\right)\\ =\frac{2}{2}\\ =1\end{array}$

  $\begin{array}{c}{a}_7=3a_6+1\left(\because a_6=1\text{ is odd}\right)\\ =3\left(1\right)+1\\ =4\end{array}$

  $\begin{array}{c}{a}_8=\frac{a_7}{2}\left(\because a_7=4\text{ is even}\right)\\ =\frac{4}{2}\\ =2\end{array}$

  $\begin{array}{c}{a}_9=\frac{a_8}{2}\left(\because a_8=2\text{ is even}\right)\\ =\frac{2}{2}\\ =1\end{array}$

  $\begin{array}{c}{a}_{10}=3a_9+1\left(\because a_9=1\text{ is odd}\right)\\ =3\left(1\right)+1\\ =4\end{array}$

The obtained sequence is 5, 16, 8, 4, 2, 1, 4, 2, 1, 4.

To determine the first 10 terms of the sequence for which $a_1=12$ , put $n=2,\dots ,10$ as shown below:

  $\begin{array}{c}{a}_2=\frac{a_1}{2}\left(\because a_1=12\text{ is even}\right)\\ =\frac{12}{2}\\ =6\end{array}$

  $\begin{array}{c}{a}_3=\frac{a_2}{2}\left(\because a_2=6\text{ is even}\right)\\ =\frac{6}{2}\\ =3\end{array}$

  $\begin{array}{c}{a}_4=3a_3+1\left(\because a_3=3\text{ is odd}\right)\\ =3\left(3\right)+1\\ =10\end{array}$

  $\begin{array}{c}{a}_5=\frac{a_4}{2}\left(\because a_4=10\text{ is even}\right)\\ =\frac{10}{2}\\ =5\end{array}$

  $\begin{array}{c}{a}_6=3a_5+1\left(\because a_5=5\text{ is odd}\right)\\ =3\left(5\right)+1\\ =16\end{array}$

  $\begin{array}{c}{a}_7=\frac{a_6}{2}\left(\because a_6=16\text{ is even}\right)\\ =\frac{16}{2}\\ =8\end{array}$

  $\begin{array}{c}{a}_8=\frac{a_7}{2}\left(\because a_7=8\text{ is even}\right)\\ =\frac{8}{2}\\ =4\end{array}$

  $\begin{array}{c}{a}_9=\frac{a_8}{2}\left(\because a_8=4\text{ is even}\right)\\ =\frac{4}{2}\\ =2\end{array}$

  $\begin{array}{c}{a}_{10}=\frac{a_9}{2}\left(\because a_8=2\text{ is even}\right)\\ =\frac{2}{2}\\ =1\end{array}$

The obtained sequence is 12, 6, 3, 10, 5, 16, 8, 4, 2, 1.

From the three sequence, it can be observed that all of them end in 4, 2, 1 with these 3 numbers on repeat in the same order.