n?1n所以?Tn?1?20??2?1??21??3?2??22?????? ?n?n?1?2?n?2?????Tn?2?2?2?????2012n?1?n?2?n11?2n???n?2??1?n??21?2nn?1,故Tn??n?1??2n?1,所以
nn?Tn?1?n?2n?1?1,所以Tn?1?Tn?n?2n?1?1??n?1?2?1?n?1?2?0,所以Tn?1?Tn,所以?????????Tn?是递增数列,所以?Tn?min?T1?1,所以m?1,所以m的最大值为1
20.【安徽省淮南市2018届第四次联考】已知数列?an?的前n项和为Sn,且对任意正整数n,都有
4an?3Sn?2成立.记bn?log2an.
(Ⅰ)求数列?an?和?bn?的通项公式; (Ⅱ)设cn?413,数列?cn?的前n项和为Tn,求证: ?Tn?.
34?bn?1???bn?1?3?
9
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