# Chapter 7 - Section 7.1 - Sequences and Series - 7.1 Exercises: 8

The first five terms are: $2, -3, 4, -5, \text{ and }6$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the first five terms of the given sequence, $a_n=(-1)^{n-1}(n+1) ,$ substitute $n$ with the numbers from $1$ to $5.$ $\bf{\text{Solution Details:}}$ Substituting $n$ with $1$, then the first term is \begin{array}{l}\require{cancel} a_1=(-1)^{1-1}(1+1) \\\\ a_1=(-1)^{0}(2) \\\\ a_1=(1)(2) \\\\ a_1=2 .\end{array} Substituting $n$ with $2$, then the second term is \begin{array}{l}\require{cancel} a_2=(-1)^{2-1}(2+1) \\\\ a_2=(-1)^{1}(3) \\\\ a_2=(-1)(3) \\\\ a_2=-3 .\end{array} Substituting $n$ with $3$, then the third term is \begin{array}{l}\require{cancel} a_3=(-1)^{3-1}(3+1) \\\\ a_3=(-1)^{2}(4) \\\\ a_3=(1)(4) \\\\ a_3=4 .\end{array} Substituting $n$ with $4$, then the fourth term is \begin{array}{l}\require{cancel} a_4=(-1)^{4-1}(4+1) \\\\ a_4=(-1)^{3}(5) \\\\ a_4=(-1)(5) \\\\ a_4=-5 .\end{array} Substituting $n$ with $5$, then the fifth term is \begin{array}{l}\require{cancel} a_5=(-1)^{5-1}(5+1) \\\\ a_5=(-1)^{4}(6) \\\\ a_5=(1)(6) \\\\ a_5=6 .\end{array} Hence, the first five terms are: $2, -3, 4, -5, \text{ and }6$

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