Chapter 9.4, Problem 64E

To determine

To find: The first six terms of the sequence $a_1=3,a_n=2a_{n-1}$ beginning with the term $a_1$ and the first and second differences of the sequence . Also, state whether the sequence has a perfect linear model, a perfect square model, or neither.

Answer to Problem 64E

The first 6 terms of the sequence are

  $a_1=3,a_2=6,a_3=12,a_4=24,a_5=48,a_6=96$

The first differences are 3,6,12,24,48

The second differences are 3,6,12,24

The sequence has neither perfectly linear model nor perfectly quadratic model.

Explanation of Solution

Given:

  $a_1=3,a_n=2a_{n-1}$

Calculation:

The first term of the sequence is $a_1=3$

The recursive relation for the nth term of the sequence is

  $a_n=2a_{n-1}$

By substituting $n=2,3,4,5,6$ , we get

  $\begin{array}{l}{a}_2=2a_1=2\left(3\right)=6\\ a_3=2a_2=2\left(6\right)=12\\ a_4=2a_3=2\left(12\right)=24\\ a_5=2a_4=2\left(24\right)=48\\ a_6=2a_5=2\left(48\right)=96\end{array}$

The first differences of the sequence are

  $\begin{array}{l}{x}_1=a_2-a_1=6-3=3\\ x_2=a_3-a_2=12-6=6\\ x_3=a_4-a_3=24-12=12\\ x_4=a_5-a_4=48-24=24\\ x_5=a_6-a_5=96-48=48\end{array}$

The second differences of the sequence are

  $\begin{array}{l}{x}_2-x_1=6-3=3\\ x_3-x_2=12-6=6\\ x_4-x_3=24-12=12\\ x_5-x_4=48-24=24\end{array}$

Since, the neither second difference nor first difference between any two consecutive terms is constant, therefore, the sequence has neither perfectly linear model nor perfectly quadratic model.