Sn为等差数列{an}的前n项和,若a2nan=4n−12n−1,则S2nSn=_.

问题描述:

Sn为等差数列{an}的前n项和,若

a2n
an
4n−1
2n−1
,则
S2n
Sn
=______.

解析:答  由

a2n
an
4n−1
2n−1

即 
an+nd
an
4n−1
2n−1
,得an
2n−1
2
d,a1
d
2

Sn
n(a1+an)
2
n2d
2
S2n
(2n)2d
2
=4Sn

S2n
Sn
=4.
故答案为4.