Question

DISCUSS: A Recursively Defined Sequence Find the first 40 terms of the sequence defined by

$a_{n+1}=\left\{\begin{array}{ll}\frac{a_n}{2}& \text{if}{a}_n\text{isanevennumber}\\ 3a_n+1& \text{if}{a}_n\text{isanoddnumber}\end{array}\right.$

and $a_1=11$. Do the same if $a_1=25$ . Make a conjecture about this type of sequence. Try several other values for $a_1$, to test your conjecture.

Short answer

The first 40 terms of the sequence for$a_1=11$ are defined by$a_{n+1}=\frac{a_n}{2}$
are$a_{n+1}=3a_n+1$

$11,34,17,52,26,13,40,20,10,5,16,8,4,5,2,1,4,2,1,4,2,1,4,2,1,4,2,1,4,2,1,$………..znd for$a_1=25$is $25,76,38,19,58,29,88,44,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1,4,2,1,4,2,1,4,2,1..$

Step-by-step solution

Step 1.Given information.

Given sequence$a_{n+1}=\left\{\begin{array}{ll}\frac{a_n}{2}& \text{if}{a}_n\text{isanevennumber}\\ 3a_n+1& \text{if}{a}_n\text{isanoddnumber}\end{array}\right.$

Step 2. Try several other values for , to test your conjecture.

First of all make separate formula by using $a_n+1$.

For role="math" localid="1648653143070" $a_1=11$
For odd numbers use formula$a_{n+1}=3a_n+1$

$a_1=11$

Now, $a_2=3a_1+1$

$\begin{array}{c}=3\times 11+1\\ =34\end{array}$

For even numbers use the formula $a_{n+1}=\frac{a_n}{2}$

$\begin{array}{l}{a}_3=\frac{a_2}{2}\\ =\frac{34}{2}\\ =17\end{array}$

So first 40 terms are :

$11,34,17,52,26,13,40,20,10,5,16,8,4,2,1,4,2,1,4,2,1,4,2,1,4,2,1,4,2,1,\dots$

For $a_1=25$

$\begin{array}{c}{a}_2=3\times a_1+1\\ =75+1\\ =76\\ a_3=\frac{a_2}{2}\\ =\frac{76}{2}\\ =38\end{array}$

So first 40 terms are :

$25,76,38,19,58,29,88,44,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1,4,2,1,4,2,1,4,2,\mathrm{1..}$

Conjecture about the sequence is that irrespective of the value of the first term the sequence continues with the alternate form of 4,2,1 values.