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