Answer
The first five terms are: $
7,
7,
9,
13, \text{ and }
19
$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the first five terms of the given sequence, $
a_n=\dfrac{n^3+27}{n+3}
,$ 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)^3+27}{1+3}
\\\\
a_1=\dfrac{1+27}{1+3}
\\\\
a_1=\dfrac{28}{4}
\\\\
a_1=7
.\end{array}
Substituting $n$ with $2$, then the second term is
\begin{array}{l}\require{cancel}
a_2=\dfrac{(2)^3+27}{2+3}
\\\\
a_2=\dfrac{8+27}{2+3}
\\\\
a_2=\dfrac{35}{5}
\\\\
a_2=7
.\end{array}
Substituting $n$ with $3$, then the third term is
\begin{array}{l}\require{cancel}
a_3=\dfrac{(3)^3+27}{3+3}
\\\\
a_3=\dfrac{27+27}{3+3}
\\\\
a_3=\dfrac{54}{6}
\\\\
a_3=9
.\end{array}
Substituting $n$ with $4$, then the fourth term is
\begin{array}{l}\require{cancel}
a_4=\dfrac{(4)^3+27}{4+3}
\\\\
a_4=\dfrac{64+27}{4+3}
\\\\
a_4=\dfrac{91}{7}
\\\\
a_4=13
.\end{array}
Substituting $n$ with $5$, then the fifth term is
\begin{array}{l}\require{cancel}
a_5=\dfrac{(5)^3+27}{5+3}
\\\\
a_5=\dfrac{125+27}{8}
\\\\
a_5=\dfrac{152}{8}
\\\\
a_5=19
.\end{array}
Hence, the first five terms are $
7,
7,
9,
13, \text{ and }
19
.$
