Answer
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:
$2, -3, 4, -5, \text{ and }6$
