Chapter 8.2, Problem 18E

To determine

To identify: Whether the sequence $a_n={\left(-1\right)}^{2n+1}$ is arithmetic or not.

Answer to Problem 18E

The sequence $a_n={\left(-1\right)}^{2n+1}$ is arithmetic because the difference between each two term is common.

Explanation of Solution

Given:

The sequence $a_n={\left(-1\right)}^{2n+1}$

Concept used:

A sequence is arithmetic when the differences between consecutive terms are the same. So, the sequence

  $a_1,a_2,a_3,\mathrm{....},a_n,\mathrm{.....}$ is arithmetic when there is a number $d$ such that

  $a_2-a_1=a_3-a_2=a_4-a_3=a_{n+1}-a_n=\mathrm{.....}=d$

The number $d$ is called the common difference of the arithmetic sequence.

Calculation:

First calculate the first five terms of the sequence.

Calculate the 1^st term $a_1$ by substituting $n=1$ in $a_n={\left(-1\right)}^{2n+1}$ .

  $\begin{array}{c}{a}_1={\left(-1\right)}^{2\left(1\right)+1}\\ ={\left(-1\right)}^{2+1}\\ ={\left(-1\right)}^3\\ =-1\end{array}$

Calculate the 2^nd term $a_2$ by substituting $n=2$ in $a_n={\left(-1\right)}^{2n+1}$ .

  $\begin{array}{c}{a}_2={\left(-1\right)}^{2\left(2\right)+1}\\ ={\left(-1\right)}^{4+1}\\ ={\left(-1\right)}^5\\ =-1\end{array}$

Calculate the 3^rd term $a_3$ by substituting $n=3$ in $a_n={\left(-1\right)}^{2n+1}$ .

  $\begin{array}{c}{a}_3={\left(-1\right)}^{2\left(3\right)+1}\\ ={\left(-1\right)}^{6+1}\\ ={\left(-1\right)}^7\\ =-1\end{array}$

Calculate the 4^th term $a_4$ by substituting $n=4$ in $a_n={\left(-1\right)}^{2n+1}$ .

  $\begin{array}{c}{a}_4={\left(-1\right)}^{2\left(4\right)+1}\\ ={\left(-1\right)}^{8+1}\\ ={\left(-1\right)}^9\\ =-1\end{array}$

Calculate the 5^th term $a_5$ by substituting $n=5$ in $a_n={\left(-1\right)}^{2n+1}$ .

  $\begin{array}{c}{a}_5={\left(-1\right)}^{2\left(5\right)+1}\\ ={\left(-1\right)}^{10+1}\\ ={\left(-1\right)}^{11}\\ =-1\end{array}$

Hence, the sequence obtained from $a_n={\left(-1\right)}^{2n+1}$ is $-1,-1,-1,-1,-1,\mathrm{...}$ .

Calculate the difference between $a_2$ and $a_1$ .

  $\begin{array}{c}{d}_1=a_2-a_1\\ =-1-\left(-1\right)\\ =-1+1\\ =0\end{array}$

Since all the terms in the sequence being the same, the common difference $d_1=d_2=d_3=d_4=0$ .

Thus, from the above calculation the difference between each two term is common. So the sequence is arithmetic.