等差数列{an}，{bn}的前n项和分别为Sn，Tn，若SnTn=2n3n+1，则anbn=．

Question

等差数列{a_n}，{b_n}的前n项和分别为S_n，T_n，若(Sn)/(Tn)=(2n)/(3n+1)，则(an)/(bn)=______．

Answer 1

∵(Sn)/(Tn)=(2n)/(3n+1)，
∴(S2n−1)/(T2n−1)＝(2(2n−1))/(3(2n−1)+1)＝(4n−2)/(6n−2)＝(2n−1)/(3n−1)．
∴(an)/(bn)＝((2n−1)an)/((2n−1)bn)＝(S2n−1)/(T2n−1)＝(2n−1)/(3n−1)．
故答案为：(2n−1)/(3n−1)．