• 数列{an}中,a1=12, an+1=nan(n+1)(nan+1)(n∈N*),其前n项的和为Sn.求证:nΣi=1(1-SiSi+1)1√Si+1<2(√2-1).试题及答案-解答题-云返教育

    • 试题详情

      数列{an}中,a1=
      1
      2
      , an+1=
      nan
      (n+1)(nan+1)
      (n∈N*),其前n项的和为Sn
      求证:
      nΣi=1(1-
      Si
      Si+1
      )
      1
      Si+1
      <2(
      2
      -1).

      试题解答


      见解析
      证明:假设bn=
      1
      nan
      ,∴bn+1=
      1
      (n+1)an+1

      an+1=
      nan
      (n+1)(nan+1)

      bn+1-bn=
      1
      (n+1)an+1
      -
      1
      nan
      =
      1
      (n+1)
      nan
      (n+1)(nan+1)
      -
      1
      nan
      =
      nan+1
      nan
      -
      1
      nan
      =1(3分)
      ∴{b
      n}是首项为2,公差为1的等差数列.(4分)
      ∵b
      n=2+(n-1)?1=n+1,∴an=
      1
      nbn
      =
      1
      n(n+1)
      =
      1
      n
      -
      1
      n+1
      ,(6分)
      Sn=(1-
      1
      2
      )+(
      1
      2
      -
      1
      3
      )++(
      1
      n
      -
      1
      n+1
      )=1-
      1
      n+1
      =
      n
      n+1

      Si
      Si+1
      =
      i(i+2)
      (i+1)2
      =
      i2+2i
      i2+2i+1
      <1∴(1-
      Si
      Si+1
      )
      1
      Si+1
      =(
      1
      Si
      -
      1
      Si+1
      )
      Si
      Si+1
      =(
      1
      Si
      -
      1
      Si+1
      )(
      1
      Si
      +
      1
      Si+1
      )
      Si
      Si+1

      =(
      1
      Si
      -
      1
      Si+1
      )(
      Si
      Si+1
      +
      Si
      Si+1
      )<2(
      1
      Si
      -
      1
      Si+1
      )
      nΣi=1(1-
      Si
      Si+1
      )
      1
      Si+1
      <2[(
      1
      S1
      -
      1
      S2
      )+(
      1
      S2
      -
      1
      S3
      )+(
      1
      Sn
      -
      1
      Sn+1
      )]=2(
      1
      S1
      -
      1
      Sn+1
      )=2(
      2
      -
      n+2
      n+1
      )<2(
      2
      -1).(15分)
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