数列1,2+12,3+12+14,…,n+12+14+…+1-查字典问答网
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来自孙崇峰的问题

  数列1,2+12,3+12+14,…,n+12+14+…+12n−1的前n项和为()A.n+1-(12)n−1B.12n2+32n+12n−1-3C.12n2+32n+12n−1-2D.n+12n−1-1

  数列1,2+12,3+12+14,…,n+12+14+…+12n−1的前n项和为()

  A.n+1-(12)n−1

  B.12n2+32n+12n−1-3

  C.12n2+32n+12n−1-2

  D.n+12n−1-1

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2020-11-1718:42
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陈志坚

  ∵此数列的通项an=n+12+14

2020-11-17 18:43:47
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