问题 选择题
等差数列{an},{bn}的前n项和分别为Sn,Tn,若
Sn
Tn
=
2n
3n+1
,则
an
bn
=(  )
A.
2
3
B.
2n-1
3n-1
C.
2n+1
3n+1
D.
2n-1
3n+4
答案

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.

选择题
单项选择题