已知数列{a n }是等差数列,a 2 =6,a 5 =12,数列{b n }的前n项和是S n ,且S n + b n =1. (1)求数列{a n }的通项公式. (2)求证:数列{b n }是等比数列. (3)记c n = ,{c n }的前n项和为T n ,若T n $amp; 对一切n∈N * 都成立,求最整数m.
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