题目

已知数列{log2(an-1)}(n∈N*)为等差数列,且a1=3,a3=9. (1)求数列{an}的通项公式; (2)证明:+…+<1. 答案：(1)解设等差数列{log2(an-1)}的公差为d.由a1=3,a3=9,得log22+2d=log28,即d=1. ∴log2(an-1)=1+(n-1)×1=n,即an=2n+1. (2)证明, +…+ =+…+ ==1-<1.

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