College Algebra (11th Edition)

The first five terms are: $0, \dfrac{1}{9}, \dfrac{2}{27}, \dfrac{1}{27}, \text{ and } \dfrac{4}{243}$
$\bf{\text{Solution Outline:}}$ 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 $0, \dfrac{1}{9}, \dfrac{2}{27}, \dfrac{1}{27}, \text{ and } \dfrac{4}{243} .$