Question

已知数列{an}中,a1=1,前n项和Sn=an.

(1)求a2,a3;

(2)求{an}的通项公式.

Answer 1

解:(1)由S2=a2得3(a1+a2)=4a2,解得a2=3a1=3,

由S3=a3得3(a1+a2+a3)=5a3,

解得a3= (a1+a2)=6.

(2)由题设知a1=1.

当n>1时有an=Sn-Sn-1=an-an-1,

整理得an=an-1,

于是a1=1,

a2=a1,

a3=a2,

…

an-1=an-2,

an=an-1.

将以上n个等式两端分别相乘,整理得an=.

综上,{an}的通项公式an=.