Chapter 8.1, Problem 44E

To determine

To Compute : The first 40 terms of the sequence when $a_1=11\text{and}{a}_1=25$ .

Answer to Problem 44E

When $a_1=11$

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

When the first 40 terms are : 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, …., 4, 2, 1, 4 .

Explanation of Solution

Given information :

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

Considering, $a_1=11$

Gives, $a_2=3\times 11+1=34$

  $a_3=\frac{1}{2}\times 34=17$

  $a_4=3\times 17+1=52$ etc . . .

We can find all the 40 terms similarly.

Thus, the first 40 terms of the sequence when $a_1=11$ is : 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, ..., 4, 2, 1, 4 .

Considering, $a_1=25$

Gives, $a_2=3\times 25+1=76$

  $a_3=\frac{1}{2}\times 76=38$

  $a_4=\frac{1}{2}\times 38=17$ etc . . .

Thus, the first 40 terms of the sequence when $a_1=25$

is : 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, ...., 4, 2, 1, 4 .

As $a_1$ is an odd number we get $a_2$ as an even number, because $a_2=3a_1+1$ . Similarly, for $a_n$ being an odd number, $a_{n+1}=3a_n+1$ will always be an even number therefore, $a_{n+2}=\frac{a_{n+1}}{2}=\frac{3a_n+1}{2}$ . The sequence then starts reducing the odd factors until it’s greatest odd factor is1, then the terms just starts oscillating between 4, 2, 1 .

Thus, we can conclude that

Irrespective of $a_1$ these obtained sequences are eventually reduced to having 1 as their greatest odd factor.