数列an中已知a1=1,a2=5,a(n+2)=a(n+1)-an,则a2009等于

问题描述:

数列an中已知a1=1,a2=5,a(n+2)=a(n+1)-an,则a2009等于

a(n+3)=a(n+2)-a(n+1)=a(n+1)-an-a(n+1)=-an
因此a是以1,5,4,-1,-5,-4做周期为6的循环
2008除以6余数是5
则a2008=a5=-5

a(n+3)=a(n+2)-a(n+1)=-an
a2009=a5=-a2=-5