Question

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

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{an}的通项公式

Answer 1

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

Answer 2

两式相减可得a(n+1)=a*an 又s1=a1=a（sn-an+1）=a(a1-a1+1)=a
所以an为等比数列，an=a的n次方。