已知：an+sn=n.1、令bn=an-1,求证：{bn}是等比数列.2、求an

Sn+an=nS(n-1)+a(n-1)=n-1an+an-a(n-1)=12an=a(n-1)+1bn=an-12an-2=a(n-1)-12bn=b(n-1)bn=(1/2)b(n-1)故等比a1+a1=1a1=1/2b1=a1-1=-1/2bn=(1/2)^(n-1)*b1=-(1/2)^n=an-1an=1-(1/2)^n