问题
填空题
Sn为等差数列{an}的前n项和,若
|
答案
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.
Sn为等差数列{an}的前n项和,若
|
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.