College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 7 - Section 7.1 - Sequences and Series - 7.1 Exercises: 6

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} .$
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