Answer
The first five terms are: $
1,
\dfrac{7}{6},
1,
\dfrac{5}{6}, \text{ and }
\dfrac{19}{27}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the first five terms of the given sequence, $
a_n=\dfrac{4n-1}{n^2+2}
,$ 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{4(1)-1}{(1)^2+2}
\\\\
a_1=\dfrac{4-1}{1+2}
\\\\
a_1=\dfrac{3}{3}
\\\\
a_1=1
.\end{array}
Substituting $n$ with $2$, then the second term is
\begin{array}{l}\require{cancel}
a_2=\dfrac{4(2)-1}{(2)^2+2}
\\\\
a_2=\dfrac{8-1}{4+2}
\\\\
a_2=\dfrac{7}{6}
.\end{array}
Substituting $n$ with $3$, then the third term is
\begin{array}{l}\require{cancel}
a_3=\dfrac{4(3)-1}{(3)^2+2}
\\\\
a_3=\dfrac{12-1}{9+2}
\\\\
a_3=\dfrac{11}{11}
\\\\
a_3=1
.\end{array}
Substituting $n$ with $4$, then the fourth term is
\begin{array}{l}\require{cancel}
a_4=\dfrac{4(4)-1}{(4)^2+2}
\\\\
a_4=\dfrac{16-1}{16+2}
\\\\
a_4=\dfrac{15}{18}
\\\\
a_4=\dfrac{5}{6}
.\end{array}
Substituting $n$ with $5$, then the fifth term is
\begin{array}{l}\require{cancel}
a_5=\dfrac{4(5)-1}{(5)^2+2}
\\\\
a_5=\dfrac{20-1}{25+2}
\\\\
a_5=\dfrac{19}{27}
.\end{array}
Hence, the first five terms are $1,
\dfrac{7}{6},
1,
\dfrac{5}{6}, \text{ and }
\dfrac{19}{27}$
