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已知数列{an}满足an+1=an﹣an﹣1(n≥2),a1=m,a2=n,Sn为数列{an}的前n项和,则S...
  • 已知数列{an}满足an+1=an﹣an﹣1(n≥2),a1=m,a2=n,Sn为数列{an}的前n项和,则S...

  • 问题详情:已知数列{an}满足an+1=an﹣an﹣1(n≥2),a1=m,a2=n,Sn为数列{an}的前n项和,则S2017的值为()A.2017n﹣m    B.n﹣2017m    C.m    D.n【回答】C【考点】数列递推式.【分析】an+1=an﹣an﹣1(n≥2),a1=m,a2=n,可得an+6=an.即可得出.【解答】解:∵an+1=an﹣an﹣1(n≥2),a1=m,a2=n,∴a3=n﹣m,a4=﹣m,a5=﹣n,a6=...
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