根号下Sn是1/4与（an+1)的平方的等比中项,则：Sn=(an+1)^2/4 an=Sn-S=(an+1)^2/4-(a+1)^2/4 4an=an^2+2an-a^2-2a an^2-2an+1=a^2+2a+1 (an-1)^2=(a+1)^2 由于数列为正项数列 故：an-1=a+1 an-a=2 所以数列{an}是等差数列,且公差d=2 Sn=(an+1)^2/4 故a1=S1=(a1+1)^2/4 a1^2+2a1+1=4a1 (a1-1)^2=0 a1=1 an=a1+(n-1)d=1+2(n-1)=2n-1 bn=an/2^n=(2n-1)/2^n Tn不太好求.