问题 填空题
Sn为等差数列{an}的前n项和,若
a2n
an
=
4n-1
2n-1
,则
S2n
Sn
=______.
答案

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.

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