两等差数列{an}和{bn},前n项和分别为Sn,Tn,且SnTn=7n+2/n+3,则a2+a20b7+b15=_.
问题描述:
两等差数列{an}和{bn},前n项和分别为Sn,Tn,且
=Sn Tn
,则7n+2 n+3
=______.
a2+a20
b7+b15
答
在{an}为等差数列中,当m+n=p+q(m,n,p,q∈N+)时,am+an=ap+aq.
所以
=
a2+a20
b7+b15
=21×(a1+a21)×
1 2 21×(b1+b21)×
1 2
,S21 T21
又因为
=Sn Tn
,7n+2 n+3
所以
=
a2+a20
b7+b15
.149 24
故答案为:
.149 24