摘要：(1)解:设数列{bn}的公差为d,由题意得,∴bn=3n-2 (2)证明:由bn=3n-2知 Sn=loga(1+1)+loga(1+)+-+loga(1+) =loga［(1+1)(1+)-(1+ )］ 而logabn+1=loga,于是.比较Sn与logabn+1?的大小比较(1+1)(1+)-(1+)与的大小. 取n=1.有(1+1)= 取n=2.有(1+1)(1+ 推测:(1+1)(1+)-(1+)＞ (*) ①当n=1时.已验证(*)式成立. ②假设n=k(k≥1)时(*)式成立.即(1+1)(1+)-(1+)＞ 则当n=k+1时. ,即当n=k+1时.(*)式成立,由①②知.(*)式对任意正整数n都成立.于是.当a＞1时.Sn＞logabn+1?,当 0＜a＜1时.Sn＜logabn+1?

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