• 已知 a1=3,a2=6,且 an+2=an+1-an,则a2009= .试题及答案-填空题-云返教育

    • 试题详情

      已知 a1=3,a2=6,且 an+2=an+1-an,则a2009=         

      试题解答


      -6
      解:由条件an+2=an+1-an可得:an+6=an+5-an+4
      =(a
      n+4-an+3)-an+4=-an+3=-(an+2-an+1
      =-[(a
      n+1-an)-an+1]=an
      于是可知数列{a
      n}的周期为6,
      ∴a
      2009=a5,又a1=3,a2=6,
      ∴a
      3=a2-a1=3,a4=a3-a2=-3,
      故a
      2009=a5=a4-a3=-6.
      故答案为:-6.
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