## Calculus: Early Transcendentals 8th Edition

$2,1,-1,-2,-1,1...$
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 $2,1,-1,-2,-1$.