Chapter 9.4, Problem 67E

To determine

To find: The first six terms of the sequence $a_1=5,a_n=4n-a_{n-1}$ 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 67E

The first 6 terms of the sequence are

  $a_1=5,a_2=3,a_3=9,a_4=7,a_5=13,a_6=11$

The first differences are -2,6,-2,6,-2

The second differences are 8,-8,8,-8

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

Explanation of Solution

Given:

  $a_1=5,a_n=4n-a_{n-1}$

Calculation:

The first term of the sequence is $a_1=5$

The recursive relation for the nth term of the sequence is

  $a_n=4n-a_{n-1}$

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

  $\begin{array}{l}{a}_2=4\left(2\right)-a_1=8-5=3\\ a_3=4\left(3\right)-a_2=12-3=9\\ a_4=4\left(4\right)-a_3=16-9=7\\ a_5=4\left(5\right)-a_4=20-7=13\\ a_6=4\left(6\right)-a_5=24-13=11\end{array}$

The first differences of the sequence are

  $\begin{array}{l}{x}_1=a_2-a_1=3-5=-2\\ x_2=a_3-a_2=9-3=6\\ x_3=a_4-a_3=7-9=-2\\ x_4=a_5-a_4=13-7=6\\ x_5=a_6-a_5=11-13=-2\end{array}$

The second differences of the sequence are

  $\begin{array}{l}{x}_2-x_1=6-\left(-2\right)=8\\ x_3-x_2=-2-6=-8\\ x_4-x_3=6-\left(-2\right)=8\\ x_5-x_4=-2-6=-8\end{array}$

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.