• 若数列{an}的前n项和Sn=2n2-2n,则数列{an}是试题及答案-单选题-云返教育

    • 试题详情

      若数列{an}的前n项和Sn=2n2-2n,则数列{an}是         

      试题解答


      A
      ∵Sn=2n2-2n,
      则S
      n-Sn-1=an=2n2-2n-[2(n-1)2-2(n-1)]=4n-4
      故数列{a
      n}是公差为4的等差数列
      故选A.

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