等比数列{An}的前项和为Sn,已知对任意的n∈N*,点（n,Sn）均在函数y=Bx（上标）+r的图像上

1)
n=1,S1=b+r=a1
n=2,S2=b^2+r,a2=a1q=S2-S1=b(b-1),q=b(b-1)/(b+r)
n=3,S3=b^3+r,a3=a1q^2=b^2(b-1)^2/(b+r)=S3-S2=b(b-1)
r=-1,q=B,a1=B-1
2)
B=2,a1=1,q=2,An=q^(n-1)=2^(n-1)
Bn=(n+1)/4*2^(n-1)=(n+1)/2^(n+1),B1=1/2
Tn=1/2+3/2^3+4/2^4+...+n/2^n+(n+1)/2^(n+1) (1)
1/2Tn=1/4+3/2^4+...+n/2^(n+1)+(n+1)/2^(n+2) (2)
(1)-(2)得
1/2Tn=5/8+1/2^4+...+1/2^(n+1)-(n+1)/2^(n+2)
=3/2-(n+3)/2^(n+1)