Chapter 9.4, Problem 65E

To determine

To find :The first six terms of the sequence 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 65E

The first 6 terms of the sequence are

  $a_1=3,a_2=7,a_3=16,a_4=32,a_5=57,a_6=93$

The first differences are 4,9,16,25,36

The second differences are 5,7,9,11

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

Explanation of Solution

Given sequence: $a_1=3,a_n=2a_{n-1}$

Calculation:

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

The recursive relation for the nth term of the sequence is

  $a_n=a_{n-1}+n^2$

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

  $\begin{array}{l}{a}_2=a_1+2^2=3+4=7\\ a_3=a_2+3^2=7+9=16\\ a_4=a_3+4^2=16+16=32\\ a_5=a_4+5^2=32+25=57\\ a_6=a_5+6^2=57+36=93\end{array}$

The first differences of the sequence are

  $\begin{array}{l}{x}_1=a_2-a_1=7-3=4\\ x_2=a_3-a_2=16-7=9\\ x_3=a_4-a_3=32-16=16\\ x_4=a_5-a_4=57-32=25\\ x_5=a_6-a_5=93-57=36\end{array}$

The second differences of the sequence are

  $\begin{array}{l}{x}_2-x_1=9-4=5\\ x_3-x_2=16-9=7\\ x_4-x_3=25-16=9\\ x_5-x_4=36-25=11\end{array}$

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.