已知数列前n项和为Sn,首项为a1,且1,an,Sn为等差数列 (1)求数列{an}的通项公式(2)设Tn为数列{n/an}的前n项和,若对于一切n∈N*,总有Tn
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![已知数列前n项和为Sn,首项为a1,且1,an,Sn为等差数列 (1)求数列{an}的通项公式(2)设Tn为数列{n/an}的前n项和,若对于一切n∈N*,总有Tn](/uploads/image/z/1087113-57-3.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E9%A6%96%E9%A1%B9%E4%B8%BAa1%2C%E4%B8%941%2Can%2CSn%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97+%281%29%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%282%29%E8%AE%BETn%E4%B8%BA%E6%95%B0%E5%88%97%7Bn%2Fan%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2C%E8%8B%A5%E5%AF%B9%E4%BA%8E%E4%B8%80%E5%88%87n%E2%88%88N%2A%2C%E6%80%BB%E6%9C%89Tn)
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