题目内容：

已知数列{an}的前n项和为Sn,满足an≠0,anSn+1-an+1Sn=(2的n-1次)an+1an,求Sn=2（n-1次）an设bn=an/an+1
求bn的前n项和Tn

优质解答

(1)anS(n+1)-a(n+1)Sn=2^(n-1)a(n+1)an两边同除a(n+1)an得：S(n+1)/a(n+1)-Sn/an=2^(n-1)设cn=Sn/an∴c(n+1)=cn+2^(n-1)∴c(n+1)-2^n=cn-2^(n-1)=...=c1-2^0=S1/a1-1=0∴cn=2^(n-1)∴Sn/an=2^(n-1)∴Sn=2^(n-1)an(2)...