数列{an}的各项均为正数,Sn为其前n项和,对于任意n∈N*,总有an,Sn,an2成等差数列.
(1)求数列{an}的通项公式;
(2)设bn=1an2,数列{bn}的前n项和为Tn,求证:Tn>nn+1.
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