## College Algebra (11th Edition)

Published by Pearson

# Chapter 7 - Section 7.1 - Sequences and Series - 7.1 Exercises - Page 635: 28

#### Answer

$a_1=-3, a_2=-12, a_3=-72, a_4=-576$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the first four terms of the given sequence \begin{array}{l}\require{cancel} a_1=-3 \\\\ a_n=2n\cdot a_{n-1} ,\text{ if } n\gt1 ,\end{array} use the given value of $a_1$ and then use substitution for the next $3$ terms. $\bf{\text{Solution Details:}}$ Using the value of $a_1$ and $n=2,$ the second term is \begin{array}{l}\require{cancel} a_2=2(2)\cdot a_{2-1} \\\\ a_2=4\cdot a_{1} \\\\ a_2=4(-3) \\\\ a_2=-12 .\end{array} Using the value of $a_2$ and $n=3,$ the third term is \begin{array}{l}\require{cancel} a_3=2(3)\cdot a_{3-1} \\\\ a_3=6\cdot a_{2} \\\\ a_3=6(-12) \\\\ a_3=-72 .\end{array} Using the value of $a_3$ and $n=4,$ the fourth term is \begin{array}{l}\require{cancel} a_4=2(4)\cdot a_{4-1} \\\\ a_4=8\cdot a_{3} \\\\ a_4=8(-72) \\\\ a_4=-576 .\end{array} Hence, $a_1=-3, a_2=-12, a_3=-72, a_4=-576 .$

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