Chapter 9.4, Problem 63E

To determine

To find :The first six terms of the sequence $a_1=4,a_n=a_{n-1}+3n$ 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 63E

The first 6 terms of the sequence are

  $a_1=4,a_2=10,a_3=19,a_4=31,a_5=46,a_6=64$

The first differences are 6, 9,12,15,18.

The second differences are 3,3,3,3.

The sequence has perfectly quadratic model.

Explanation of Solution

Given sequence: $a_1=4,a_n=a_{n-1}+3n$

Calculation:

The first term of the sequence is $a_1=4$

The recursive relation for the nth term of the sequence is

  $a_n=a_{n-1}+3n$

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

  $\begin{array}{l}{a}_2=a_1+3\left(2\right)=4+6=10\\ a_3=a_2+3\left(3\right)=10+9=19\\ a_4=a_3+3\left(4\right)=19+12=31\\ a_5=a_4+3\left(5\right)=31+15=46\\ a_6=a_5+3\left(6\right)=46+18=64\end{array}$

The first differences of the sequence are

  $\begin{array}{l}{x}_1=a_2-a_1=10-4=6\\ x_2=a_3-a_2=19-10=9\\ x_3=a_4-a_3=31-19=12\\ x_4=a_5-a_4=46-31=15\\ x_5=a_6-a_5=64-46=18\end{array}$

The second differences of the sequence are

  $\begin{array}{l}{x}_2-x_1=9-6=3\\ x_3-x_2=12-9=3\\ x_4-x_3=15-12=3\\ x_5-x_4=18-15=3\end{array}$

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