{an}满足an+1=an-an-1(n≥2),a1=a,a2=b,记Sn=a1+a2+a3+…+an,则 a2008=?S2008=?
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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=?](/uploads/image/z/15235028-44-8.jpg?t=%7Ban%7D%E6%BB%A1%E8%B6%B3an%2B1%3Dan-an-1%28n%E2%89%A52%29%2Ca1%3Da%2Ca2%3Db%2C%E8%AE%B0Sn%3Da1%2Ba2%2Ba3%2B%E2%80%A6%2Ban%2C%E5%88%99+a2008%3D%3FS2008%3D%3F)
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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=?
{an}满足an+1=an-an-1(n≥2),a1=a,a2=b,记Sn=a1+a2+a3+…+an,则 a2008=?S2008=?
{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
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
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
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