已知数列{an}的前n项和Sn满足:Sn=a(Sn-an+1) (a为常数,a不等于0,a不等于1)(1)求{a

Sn=a(Sn-an+1) =a(Sn-1+1)=aSn-1+a Sn-1=aSn-2+a Sn-Sn-1=a(Sn-1-Sn-2) an=aan-1 所以an是公比为a的等比数列 a1=a 所以an=a^n