问题 选择题
已知两个等差数列{an}和{bn}的前n项和分别An和Bn,且
An
Bn
=
7n+45
n+3
,则使得
an
bn
为整数的正整数n的值是(  )
A.1,3,5,8,11B.所有正整数C.1,2,3,4,5D.1,2,3,5,11
答案

由题意可得

an
bn
=
(2n-1)an
(2n-1)bn
=
(2n-1)
a1+a2n-1
2
(2n-1)
b1+b2n-1
2

=

A2n-1
B2n-1
=
7(2n-1)+45
(2n-1)+3
=
7n+19
n+1
=7+
12
n+1

经验证可知当n=1,2,3,5,11时,上式为正整数,

故选D

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