Answer
The first five terms are: $
0, \dfrac{1}{9}, \dfrac{2}{27}, \dfrac{1}{27}, \text{ and } \dfrac{4}{243}
$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the first five terms of the given sequence, $
a_n=\left( \dfrac{1}{3} \right)^n(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=\left( \dfrac{1}{3} \right)^1(1-1)
\\\\
a_1=\left( \dfrac{1}{3} \right)(0)
\\\\
a_1=0
.\end{array}
Substituting $n$ with $2$, then the second term is
\begin{array}{l}\require{cancel}
a_2=\left( \dfrac{1}{3} \right)^2(2-1)
\\\\
a_2=\left( \dfrac{1}{9} \right)(1)
\\\\
a_2=\dfrac{1}{9}
.\end{array}
Substituting $n$ with $3$, then the third term is
\begin{array}{l}\require{cancel}
a_3=\left( \dfrac{1}{3} \right)^3(3-1)
\\\\
a_3=\left( \dfrac{1}{27} \right)(2)
\\\\
a_3=\dfrac{2}{27}
.\end{array}
Substituting $n$ with $4$, then the fourth term is
\begin{array}{l}\require{cancel}
a_4=\left( \dfrac{1}{3} \right)^4(4-1)
\\\\
a_4=\left( \dfrac{1}{81} \right)(3)
\\\\
a_4=\dfrac{3}{81}
\\\\
a_4=\dfrac{1}{27}
.\end{array}
Substituting $n$ with $5$, then the fifth term is
\begin{array}{l}\require{cancel}
a_5=\left( \dfrac{1}{3} \right)^5(5-1)
\\\\
a_5=\left( \dfrac{1}{243} \right)(4)
\\\\
a_5=\dfrac{4}{243}
.\end{array}
Hence, the first five terms are $
0, \dfrac{1}{9}, \dfrac{2}{27}, \dfrac{1}{27}, \text{ and } \dfrac{4}{243}
.$
