Chapter 9, Problem 7PS

a.

To determine

Write the first $20$ terms of the sequence.

Answer to Problem 7PS

  $\text{7},\text{ 22},\text{ 11},\text{ 34},\text{ 17},\text{ 52},\text{ 26},\text{ 13},\text{ 4}0,\text{ 2}0,\text{ 1}0,\text{ 5},\text{ 16},\text{ 8},\text{ 4},\text{ 2},\text{ 1},\text{ 4},\text{ 2},\text{ 1}$

Explanation of Solution

Calculation:

Let us try to find the first $20$ terms of the sequence .

It is given that,

  $a_1=7$ and

  $a_n=a_{n-1},\text{ if }{a}_{n-1}\text{ is even}$

  $a_n=3a_{n-1}+1,\text{ if }{a}_{n-1}\text{is odd}$

The first $20$ terms of the sequence are

  $\begin{array}{l}{a}_1={\mathrm{7....1}}^{st}term\\ a_2=3a_1+1\\ a_2={\mathrm{22....2}}^{nd}term\\ a_3=\frac{a_2}{2}\\ a_3=\frac{22}{2}\\ a_3={\mathrm{11....3}}^{rd}term\end{array}$

  $\begin{array}{l}{a}_4=3a_3+1\\ a_4=3\left(11\right)+1\\ a_4={\mathrm{34....4}}^{th}term\\ a_5=\frac{a_4}{2}\\ a_5=\frac{34}{2}\\ a_5={\mathrm{17....5}}^{th}term\\ a_6=3a_5+1\\ a_6=3\left(17\right)+1\\ a_6={\mathrm{52....6}}^{th}term\\ a_7=\frac{a_6}{2}\\ a_7=\frac{52}{2}\\ a_7={\mathrm{26....7}}^{th}term\\ a_8=\frac{a_7}{2}\\ a_8=\frac{26}{2}\\ \end{array}$

  $\begin{array}{l}{a}_8={\mathrm{13....8}}^{th}term\\ a_9=3a_8+1\\ a_9=3\left(18\right)+1\\ a_9={\mathrm{40....9}}^{th}term\\ a_{10}=\frac{a_9}{2}\\ a_{10}=\frac{40}{2}\\ a_{10}={\mathrm{20....10}}^{th}term\\ a_{11}=\frac{a_{10}}{2}\\ a_{11}=\frac{20}{2}\\ a_{11}={\mathrm{10....11}}^{th}term\\ a_{12}=\frac{a_{11}}{2}\\ a_{12}=\frac{10}{2}\\ a_{12}={\mathrm{5....12}}^{th}term\end{array}$

  $\begin{array}{l}{a}_{13}=3a_{12}+1\\ a_{13}=3\left(5\right)+1\\ a_{13}={\mathrm{16....13}}^{th}term\end{array}$

  $\begin{array}{l}{a}_{13}=3a_{12}+1\\ a_{13}=3\left(5\right)+1\\ a_{13}={\mathrm{16....13}}^{th}term\\ a_{14}=\frac{a_{13}}{2}\\ a_{14}=\frac{16}{2}\\ a_{14}={\mathrm{8....14}}^{th}term\\ a_{15}=\frac{a_{14}}{2}\\ a_{15}=\frac{8}{2}\\ a_{15}={\mathrm{4....15}}^{th}term\\ a_{16}=\frac{a_{15}}{2}\\ a_{16}=\frac{4}{2}\\ a_{16}={\mathrm{2....16}}^{th}term\\ a_{17}=\frac{a_{16}}{2}\\ a_{17}=\frac{2}{2}\\ a_{17}={\mathrm{1....17}}^{th}term\\ \end{array}$

  $\begin{array}{l}{a}_{18}=3a_{17}+1\\ a_{18}=3\left(1\right)+1\\ a_{18}={\mathrm{4....18}}^{th}term\\ a_{19}=\frac{a_{18}}{2}\\ a_{19}=\frac{4}{2}\\ a_{19}={\mathrm{2....19}}^{th}term\\ a_{20}=\frac{a_{19}}{2}\\ a_{20}=\frac{2}{2}\\ a_{20}={\mathrm{1....20}}^{th}term\end{array}$

Hence, the first $20$ terms of the sequence

  $\text{7},\text{ 22},\text{ 11},\text{ 34},\text{ 17},\text{ 52},\text{ 26},\text{ 13},\text{ 4}0,\text{ 2}0,\text{ 1}0,\text{ 5},\text{ 16},\text{ 8},\text{ 4},\text{ 2},\text{ 1},\text{ 4},\text{ 2},\text{ 1}$

b.

To determine

Write the first $10$ terms of the sequence.

Answer to Problem 7PS

  $\text{4},\text{ 2},\text{ 1},\text{ 4},\text{ 2},\text{ 1},\text{ 4},\text{ 2},\text{ 1},\text{ 4}$

Explanation of Solution

Calculation:

The first $10$ terms of the sequence are

  $\begin{array}{l}{a}_1={\mathrm{4....1}}^{st}term\\ a_2=\frac{a_1}{2}\\ a_2=\frac{4}{2}\\ a_2={\mathrm{2....2}}^{nd}term\\ a_3=\frac{a_2}{2}\\ a_3=\frac{2}{2}\\ a_3={\mathrm{1....3}}^{rd}term\\ a_4=3a_3+1\\ a_4=3\left(1\right)+1\\ a_4={\mathrm{4....4}}^{th}term\\ a_5=\frac{a_4}{2}\end{array}$

  $\begin{array}{l}{a}_5=\frac{a_4}{2}\\ a_5=\frac{4}{2}\\ a_5={\mathrm{2....5}}^{th}term\\ a_6=\frac{a_5}{2}\\ a_6=\frac{2}{2}\\ a_6={\mathrm{1....6}}^{th}term\\ a_7=3a_6+1\\ a_7=3\left(1\right)+1\\ a_7={\mathrm{4....7}}^{th}term\\ a_8=\frac{a_7}{2}\\ a_8=\frac{4}{2}\\ a_8={\mathrm{2....8}}^{th}term\\ a_9=\frac{a_8}{2}\\ a_9=\frac{2}{2}\\ a_9={\mathrm{1....9}}^{th}term\end{array}$

  $\begin{array}{l}{a}_{10}=3a_9+1\\ a_{10}=3\left(1\right)+1\\ a_{10}={\mathrm{4....10}}^{th}term\end{array}$

Hence the sequence is $\text{4},\text{ 2},\text{ 1},\text{ 4},\text{ 2},\text{ 1},\text{ 4},\text{ 2},\text{ 1},\text{ 4}$