摘要:18.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. 解:(1)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)∵==. ∴++-+ =+++-+ ==1-<1.

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