等差数列{an}、{bn}的前n项和分别为Sn、Tn,且SnTn=7n+45n−3,则使得anbn为整数的正整数的n的个数是( ) A.3 B.4 C.5 D.6
问题描述:
等差数列{an}、{bn}的前n项和分别为Sn、Tn,且
=Sn Tn
,则使得7n+45 n−3
为整数的正整数的n的个数是( )an bn
A. 3
B. 4
C. 5
D. 6
答
∵等差数列{an}、{bn},
∴an=
,bn=
a1+a2n−1
2
,
b1+b2n−1
2
∴
=an bn
=nan nbn
=
n(a1+a2n−1) 2
n(b1+b2n−1) 2
,又S2n−1 T2n−1
=Sn Tn
,7n+45 n−3
∴
=an bn
=7+7(2n−1)+45 (2n−1)−3
,66 2n−4
经验证,当n=1,3,5,13,35时,
为整数,an bn
则使得
为整数的正整数的n的个数是5.an bn
故选C