等比数列｛an｝的前n项和为Sn,已知对任意的n∈N*,点(n,Sn),均在函数y=b的x次方+r(b＞0且b不等于1,

(1)
点(n,Sn),均在函数y=b^x +r
n=1
a1= b+r
Sn = b^n + r
an = Sn-S(n-1)
= (b-1)b^(n-1)
a1 = b-1 = b+r
r = -1
(2)
b=2
Sn = 2^n -1
an = 2^(n-1)
bn = (n+1)/(4an)
= (n+1) .(1/2)^(n+1)
= (1/2) [n.(1/2)^n] + (1/2)^(n+1)
Tn=b1+b2+...+bn
= (1/2)S + (1/2)(1 - (1/2)^n )
S = 1.(1/2)^1 + 2.(1/2)^2+...+n(1/2)^n (1)
(1/2)S = 1.(1/2)^2 + 2.(1/2)^3+...+n(1/2)^(n+1) (2)
(1)-(2)
(1/2)S = (1/2 +1/2^2+...+1/2^n) -n(1/2)^(n+1)
= (1-(1/2)^n) -n(1/2)^(n+1)
S = 2 - (n+2)(1/2)^n
Tn = (1/2)S + (1/2)(1 - (1/2)^n )
= 1 - (n+2)(1/2)^(n+1) + (1/2)(1 - (1/2)^n )
= 3/2 -(n+3)(1/2)^(n+1)