等差数列{an}和{bn}的前n项和分别为Sn和Tn,且SnTn=2n3n+1,则a5b5=914914

2025-01-31 22:39:02
推荐回答(1个)
回答1:

Sn
Tn
2n
3n+1
=
n(a1+an
2
n(b1+bn
2
=
a1+an
b1+bn

a5
b5
=
9
2
(a1+a9
9
2
(b1+b9
=
a1+a9
b1+b9

即当n=9时,
a5
b5
=
Sn
Tn
2n
3n+1
=
18
28
=
9
14

故答案为
9
14