a(n+1)-1=(1+an)/(3-an)-1=(2an-2)/(3-an)1/[a(n+1)-1]=(3-an)/2(an-1)=[2+(1-an)]/2(an-1)=1/(an-1)-1/2{1/(an-1)}是首项为1/(0-1)=-1公差为-1/2的等差数列1/(an-1)=-1+(-1/2)(n-1)=-(n+1)/2an-1=-2/(n+1)∴an=1-2/...