Chapter 9.4, Problem 68E

To determine

To find: The first six terms of the sequence $a_1=-2,a_n=a_{n-1}+4n$ 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 68E

The first 6 terms of the sequence are

  $a_1=-2,a_2=6,a_3=18,a_4=34,a_5=54,a_6=78$

The first differences are 8,12,16,20,24

The second differences are 4,4,4,4

The sequence has perfectly quadratic model.

Explanation of Solution

Given:

  $a_1=-2,a_n=a_{n-1}+4n$

Calculation:

The first term of the sequence is $a_1=-2$

The recursive relation for the nth term of the sequence is

  $a_n=a_{n-1}+4n$

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

  $\begin{array}{l}{a}_2=a_1+4\left(2\right)=-2+8=6\\ a_3=a_2+4\left(3\right)=6+12=18\\ a_4=a_3+4\left(4\right)=18+16=34\\ a_5=a_4+4\left(5\right)=34+20=54\\ a_6=a_5+4\left(6\right)=54+24=78\end{array}$

The first differences of the sequence are

  $\begin{array}{l}{x}_1=a_2-a_1=6-\left(-2\right)=8\\ x_2=a_3-a_2=18-6=12\\ x_3=a_4-a_3=34-18=16\\ x_4=a_5-a_4=54-34=20\\ x_5=a_6-a_5=78-54=24\end{array}$

The second differences of the sequence are

  $\begin{array}{l}{x}_2-x_1=12-8=4\\ x_3-x_2=16-12=4\\ x_4-x_3=20-16=4\\ x_5-x_4=24-20=4\end{array}$

Since, the second difference between any two consecutive terms is constant, therefore, the sequence has perfectly quadratic model.