设等差数列{an},{bn}的前n项和分别为Sn,Tn若对任意自然数n都有SnTn=2n-3/4n-3,则a9b5+b7+a3b8+b4的值为 _ .

问题描述:

设等差数列{an},{bn}的前n项和分别为Sn,Tn若对任意自然数n都有

Sn
Tn
=
2n-3
4n-3
,则
a9
b5+b7
+
a3
b8+b4
的值为 ___ .

由等差数列的性质和求和公式可得:

a9
b5+b7
+
a3
b8+b4
=
a9
b1+b11
+
a3
b1+b11

=
a3+a9
b1+b11
=
a1+a11
b1+b11
=
11(a1+a11)
2
11(b1+b11)
2

=
S11
T11
=
2×11-3
4×11-3
=
19
41

故答案为:
19
41