数列{an}前N项和Sn.3Sn =(an-1)，(n)为下标.求证{an}为等比数列

数列{an}前N项和Sn
3Sn =(an-1),(1)
当n>=2,
有:
3Sn-1=[a(n-1)-1],(2)
(1)-(2),
3an=an-an-1
an/an-1=-1/2,(n>=2)
当n=1,
3S1=3a1=a1-1
2a1=-1
a1=-1/2
所以
{an}是以a1=-1/2,-1/2为公比的等比数列

3Sn =An-1,所以3(Sn-1)=(An-1)-1,Sn-(Sn-1)=An,
所以3(Sn-Sn-1)=(An-1)-[(An-1)-1]=An-An-1
即3An=An-An-1,所以2An=-An-1,An/(An-1)=-1/2=q
所以{an}为等比数列

1. 3Sn=(an-1)
2. 3(S(n-1))=(a(n-1)-1)
1-2:
3an=an-a(n-1)
an=-1/2a(n-1)

楼主的题写的就不对吧，3Sn=(An-1)，怎么可能呢？