Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.1 - Sequences - 11.1 Exercises - Page 704: 12



Work Step by Step

Given: $a_{1}=2,a_{2}=1,a_{(n+1)}={a_{n}}-a_{n-1}$ $$a_{1}=2$$$$a_{2}=1$$ Use the given formula, $$a_{(n+1)}={a_{n}}-a_{n-1}$$$$a_{(2+1)}=a_3={a_{2}}-a_{2-1}=a_2-a_1=1-2=-1$$$$a_{(3+1)}=a_4={a_{3}}-a_{3-1}=a_3-a_2=-1-1=-2$$$$a_{(4+1)}=a_5={a_{4}}-a_{4-1}=a_4-a_3=-2+1=-1$$ Hence we see that the first five terms are $2,1,-1,-2,-1$.
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