若数列{an}满足a1=1,a2=5且an+2=an+1-an,求a2000

题目

若数列{an}满足a1=1,a2=5且an+2=an+1-an,求a2000

答案

a(n+2)=a(n+1)-an (1)
a(n+3)=a(n+2)-a(n+1) (2)
(1)+(2)
a(n+3)+a(n+2)=a(n+2)-an
a(n+3)=-an
a(n+6)=-a(n+3)=an
即数列每6项循环一次.
2000=2+6×333
a2000=a2=5