Chapter 9.4, Problem 66E

To determine

To find: The first six terms of the sequence $a_1=0,a_n=a_{n-1}-2n$ 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 66E

The first 6 terms of the sequence are

  $a_1=0,a_2=-4,a_3=-10,a_4=-18,a_5=-28,a_6=-40$

The first differences are -4,-6,-8,-10,-12.

The second differences are -2,-2,-2,-2.

The sequence has perfectly quadratic model.

Explanation of Solution

Given sequence: $a_1=0,a_n=a_{n-1}-2n$

Calculation:

The first term of the sequence is $a_1=0$

The recursive relation for the nth term of the sequence is

  $a_n=a_{n-1}-2n$

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

  $\begin{array}{l}{a}_2=a_1-2\left(2\right)=0-4=-4\\ a_3=a_2-2\left(3\right)=-4-6=-10\\ a_4=a_3-2\left(4\right)=-10-8=-18\\ a_5=a_4-2\left(5\right)=-18-10=-28\\ a_6=a_5-2\left(6\right)=-28-12=-40\end{array}$

The first differences of the sequence are

  $\begin{array}{l}{x}_1=a_2-a_1=-4-0=-4\\ x_2=a_3-a_2=-10-\left(-4\right)=-6\\ x_3=a_4-a_3=-18-\left(-10\right)=-8\\ x_4=a_5-a_4=-28-\left(-18\right)=-10\\ x_5=a_6-a_5=-40-\left(-28\right)=-12\end{array}$

The second differences of the sequence are

  $\begin{array}{l}{x}_2-x_1=-6-\left(-4\right)=-2\\ x_3-x_2=-8-\left(-6\right)=-2\\ x_4-x_3=-10-\left(-8\right)=-2\\ x_5-x_4=-12-\left(-10\right)=-2\end{array}$

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