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

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

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

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