• 数列{an}满足an+1+an=4n-3(n∈N*)(Ⅰ)若{an}是等差数列,求其通项公式;(Ⅱ)若{an}满足a1=2,Sn为{an}的前n项和,求S2n+1.试题及答案-解答题-云返教育

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      数列{an}满足an+1+an=4n-3(n∈N*
      (Ⅰ)若{a
      n}是等差数列,求其通项公式;
      (Ⅱ)若{a
      n}满足a1=2,Sn为{an}的前n项和,求S2n+1

      试题解答


      见解析
      解:( I)由题意得an+1+an=4n-3…①
      a
      n+2+an+1=4n+1…②.…(2分)
      ②-①得a
      n+2-an=4,
      ∵{a
      n}是等差数列,设公差为d,∴d=2,(4分)
      ∵a
      1+a2=1∴a1+a1+d=1,∴a1=-
      1
      2
      .(6分)
      an=2n-
      5
      2
      .(7分)
      (Ⅱ)∵a
      1=2,a1+a2=1,
      ∴a
      2=-1.(8分)
      又∵a
      n+2-an=4,
      ∴数列的奇数项与偶数项分别成等差数列,公差均为4,
      ∴a
      2n-1=4n-2,a2n=4n-5.(11分)
      S
      2n+1=(a1+a3+…+a2n+1)+(a2+a4+…+a2n)(12分)
      =(n+1)×2+
      (n+1)n
      2
      ×4+n×(-1)+
      n(n-1)
      2
      ×4
      =4n
      2+n+2.(14分)

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