Answer
The first five terms are: $
\dfrac{1}{2},
\dfrac{1}{2},
\dfrac{3}{8},
\dfrac{1}{4}, \text{ and }
\dfrac{5}{32}
$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the first five terms of the given sequence, $
a_n=\left( \dfrac{1}{2} \right)^n(n)
,$ 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}{2} \right)^1(1)
\\\\
a_1=\dfrac{1}{2}
.\end{array}
Substituting $n$ with $2$, then the second term is
\begin{array}{l}\require{cancel}
a_2=\left( \dfrac{1}{2} \right)^2(2)
\\\\
a_2=\left( \dfrac{1}{4} \right)(2)
\\\\
a_2=\dfrac{1}{2}
.\end{array}
Substituting $n$ with $3$, then the third term is
\begin{array}{l}\require{cancel}
a_3=\left( \dfrac{1}{2} \right)^3(3)
\\\\
a_3=\left( \dfrac{1}{8} \right)(3)
\\\\
a_3=\dfrac{3}{8}
.\end{array}
Substituting $n$ with $4$, then the fourth term is
\begin{array}{l}\require{cancel}
a_4=\left( \dfrac{1}{2} \right)^4(4)
\\\\
a_4=\left( \dfrac{1}{16} \right)(4)
\\\\
a_4=\dfrac{1}{4}
.\end{array}
Substituting $n$ with $5$, then the fifth term is
\begin{array}{l}\require{cancel}
a_5=\left( \dfrac{1}{2} \right)^5(5)
\\\\
a_5=\left( \dfrac{1}{32} \right)(5)
\\\\
a_5=\dfrac{5}{32}
.\end{array}
Hence, the first five terms are $
\dfrac{1}{2},
\dfrac{1}{2},
\dfrac{3}{8},
\dfrac{1}{4}, \text{ and }
\dfrac{5}{32}
.$
