Answer
The first five terms are: $
0,
\dfrac{3}{5},
\dfrac{4}{5},
\dfrac{15}{17}, \text{ and }
\dfrac{12}{13}
$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the first five terms of the given sequence, $
a_n=\dfrac{n^2-1}{n^2+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=\dfrac{(1)^2-1}{(1)^2+1}
\\\\
a_1=\dfrac{1-1}{1+1}
\\\\
a_1=\dfrac{0}{2}
\\\\
a_1=0
.\end{array}
Substituting $n$ with $2$, then the second term is
\begin{array}{l}\require{cancel}
a_2=\dfrac{(2)^2-1}{(2)^2+1}
\\\\
a_2=\dfrac{4-1}{4+1}
\\\\
a_2=\dfrac{3}{5}
.\end{array}
Substituting $n$ with $3$, then the third term is
\begin{array}{l}\require{cancel}
a_3=\dfrac{(3)^2-1}{(3)^2+1}
\\\\
a_3=\dfrac{9-1}{9+1}
\\\\
a_3=\dfrac{8}{10}
\\\\
a_3=\dfrac{4}{5}
.\end{array}
Substituting $n$ with $4$, then the fourth term is
\begin{array}{l}\require{cancel}
a_4=\dfrac{(4)^2-1}{(4)^2+1}
\\\\
a_4=\dfrac{16-1}{16+1}
\\\\
a_4=\dfrac{15}{17}
.\end{array}
Substituting $n$ with $5$, then the fifth term is
\begin{array}{l}\require{cancel}
a_5=\dfrac{(5)^2-1}{(5)^2+1}
\\\\
a_5=\dfrac{25-1}{25+1}
\\\\
a_5=\dfrac{24}{26}
\\\\
a_5=\dfrac{12}{13}
.\end{array}
Hence, the first five terms are $0,
\dfrac{3}{5},
\dfrac{4}{5},
\dfrac{15}{17}, \text{ and }
\dfrac{12}{13}
.$
