问题
选择题
设数列{an}的前n项和为sn,a1=1,an=
|
答案
∵an=
+2(n-1),sn n
∴sn-sn-1=
+2(n-1),(n≥2)sn n
整理可得,(n-1)sn-nsn-1=2n(n-1)
两边同时除以n(n-1)可得
-sn n
=2sn-1 n-1
∴数列{
}是以sn n
=1为首项,以2为公差的等差数列s1 1
∴s1+
+s2 2
+…+s3 3
-(n-1)2sn n
=n×1+
×2-(n-1)2n(n-1) 2
=n2-(n-1)2
=2n-1
由题意可得,2n-1=2013
解可得n=1007
故选A