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
,则4n−1 2n−1
=______. S2n Sn
答
解析:答 由
=a2n an
,4n−1 2n−1
即
=
an+nd an
,得an=4n−1 2n−1
d,a1=2n−1 2
.d 2
Sn=
=n(a1+an) 2
,S2n=
n2d 2
=4Sn.
(2n)2d 2
故
=4.S2n Sn
故答案为4.