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

题目简介

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

题目详情

Sn为等差数列{an}的前n项和,若
a2n
an
=
4n-1
2n-1
,则
S2n
Sn
=______.
题型:填空题难度:中档来源:不详

答案

解析:答  由
a2n
an
=class="stub"4n-1
2n-1

即 
an+nd
an
=class="stub"4n-1
2n-1
,得an=class="stub"2n-1
2
d,a1=class="stub"d
2

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

S2n
Sn
=4.
故答案为4.

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