等差数列{an}，{bn}的前n项和分别为Sn，Tn，若SnTn=2n3n+1，则anbn=（ ）A. 23B. 2n−13n−1C. 2n+13n+1D. 2n−13n+4
问题描述：
等差数列{an}，{bn}的前n项和分别为Sn，Tn，若
=Sn Tn
，则2n 3n+1
=（ ）an bn
A.
2 3
B.
2n−1 3n−1
C.
2n+1 3n+1
D.
2n−1 3n+4
答
∵
＝an bn
＝2an
2bn
=
a1+a2n−1
b1+b2n−1
＝
(2n−1)(a1+a2n−1) 2
(2n−1)(b1+b2n−1) 2
s2n−1 T2n−1
∴
=an bn
=2(2n−1) 3(2n−1)+1
2n−1 3n−1
故选B．
答案解析：利用等差数列的性质求得
＝an bn
，然后代入s2n−1 T2n−1
=Sn Tn
即可求得结果．2n 3n+1
考试点：等差数列的性质．
知识点：此题考查学生灵活运用等差数列通项公式化简求值，做题时要认真，是一道基础题．