等差数列{an}，an=2n-1，等比数列{bn}，bn=2n-1，求{anbn}的前n项和．

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等差数列{a_n}，a_n=2n-1，等比数列{b_n}，b_n=2^n-1，求{a_nb_n}的前n项和．

令T_n为{a_nb_n}的前n项和，那么：
T_n=a₁b₁+a₂b₂+…+a_nb_n
=1×2⁰+3×2¹+5×2²+…+（2n-1）•2^n-1
2T_n=1×2¹+3×2²+5×2³+…（2n-1）•2ⁿ
∴T_n=2T_n-T_n=-2（2¹+2²+…+2^n-1）+（2n-1）•2ⁿ-1×2⁰
=（2n-2）•2ⁿ+1
故答案为：（2n-2）•2ⁿ+1

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