Question

{an}满足an+1=an-an-1(n≥2),a1=a,a2=b,记Sn=a1+a2+a3+…+an,则 a2008=?S2008=?

Answer 1

利用递推关系可得
a1=a,a2=b,a3=b-a,a4=-a,a5=-b,a6=a-b,a7=a,
于是可得该数列为周期数列，周期为6，
于是a2008 = a4 = -a
S2008=S2004+a2005+a2006+a2007+a2008=0+a1+a2+a3+a4=2b-a

Answer 2

a1=a
a2=b
a3=b-a(因为a(n+1)=an-a(n-1))
a4=-a
a5=-b
a6=a-b
a7=a
出现循环
a(6m-5)=a
a(6m-4)=b
a(6m-3)=b-a
a(6m-2)=-a
a(6m-1)=-b
a(6m)=a-b
a(2008)=a(335*6-2)=-a
S2008=S(334*6+4)=334*0+a+b+(b-a)+(-a)=2b-a

Answer 3

由an+1=an-an-1得
a1=a
a2=b
a3=a2-a1=b-a
a4=a3-a2=-a
a5=a4-a3=-b
a6=a5-a4=-b+a
a7=a6-a5=a
由此可知an是以6为周期的数列
而2008/6=333
故a2008=-b+a
同理可得sn也是以6为周期
S2008=0