Chapter 8.5, Problem 64E

To determine

To write: the terms of the sequence until the discovered a pattern.

Answer to Problem 64E

Yes, the sequence has the subsequence 4,2,1 which repeated indefinitely..

Explanation of Solution

Given:

The given pattern is

  $a_{n+1}=\{\begin{array}{l}\frac{1}{2}{a}_n,\text{if }{a}_n\text{ is even}\\ 3a_n+1,\text{if }{a}_n\text{ is odd}\end{array}$

Calculation:

According to the given pattern

  $\begin{array}{l}{a}_1=34\\ a_2=\frac{1}{2}{a}_1=17\\ a_3=3a_2+1=3\left(17\right)+1=52\\ a_4=\frac{1}{2}{a}_3=26\\ a_5=\frac{1}{2}{a}_4=13\\ a_6=3a_5+1=3\left(13\right)+1=40\\ a_7=\frac{1}{2}{a}_6=20\\ a_8=\frac{1}{2}{a}_7=10\\ a_9=\frac{1}{2}{a}_8=5\\ a_{10}=3a_9+1=3\left(5\right)+1=16\end{array}$

  $\begin{array}{l}{a}_{11}=\frac{1}{2}{a}_{10}=8\\ a_{12}=\frac{1}{2}{a}_{11}=4\\ a_{13}=\frac{1}{2}{a}_{12}=2\\ a_{14}=\frac{1}{2}{a}_{13}=1\\ a_{15}=3a_{14}+1=3\left(1\right)+1=4\end{array}$

It is found that that the 4 term again and the sequence 4,2,1 will be repeated indefinitely.

Substitutes $a_1=25$ , then

  $\begin{array}{l}{a}_1=25\\ a_2=3a_1+1=3\left(25\right)+1=76\\ a_3=\frac{1}{2}{a}_2=38\\ a_4=\frac{1}{2}{a}_3=19\\ a_5=3a_4+1=3\left(19\right)+1=58\\ a_6=\frac{1}{2}{a}_5=29\\ a_7=3a_6+1=3\left(29\right)+1=88\\ a_8=\frac{1}{2}{a}_7=44\\ a_9=\frac{1}{2}{a}_8=22\\ a_{10}=\frac{1}{2}{a}_9=11\\ a_{11}=3a_{10}+1=3\left(11\right)+1=34\\ \mathrm{......}\end{array}$

From $a_{11}=34$ , then it is already in the sequence above.

It is noticed that each time an even number is reached, it is diminished until it reaches an odd number then it grows again, but at the same point it reached 4 from which the sequence has the subsequence 4,2,1 which repeated indefinitely.