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^x+r(b>0)且b≠1,b,r均为常数）的图像上.

(1)
点（n,Sn）均在函数y=b^x+r
n=1,a1=b+r (1)
n=2,
S2= b^2 +r
a2+(b+r)=b^2 +r
a2 = b(b-1) (2)
n=3,
S3 =b^3+r
a3+ b^2 +r =b^3+r
a3= b^2(b-1) (3)
a3/a2 = a2/a1
b^2(b-1)/[b(b-1)] = b(b-1)/(b+r)
b(b+r) =b(b-1)
br= -b
r= -1
(2)
b=2
Sn=2^n-1
an = Sn-S(n-1) = 2^(n-1)
bn=(n/2)an
= (1/2)(n.2^(n-1) )
consider
1+x+x^2+..+x^n = (x^(n+1)- 1)/(x-1)
1+2x+..+nx^(n-1) =[(x^(n+1)- 1)/(x-1)]'
= [nx^(n+1) - (n+1)x^n + 1]/(x-1)^2
put n=2
summation(1:1->n)i.2^(i-1)
=n.2^(n+1) - (n+1).2^n + 1
= 1+ (n-1).2^n
bn=(n/2)an
= (1/2)(n.2^(n-1) )
Tn=b1+b2+...+bn
=(1/2){summation(1:1->n)i.2^(i-1)}
=(1/2)[1+ (n-1).2^n]
(3)
b=3
Sn=3^n-1
an= Sn-S(n-1) = 2.3^(n-1)
cn = 2an/（an+1)(3an+1)
= 4.3^(n-1) /[( 1+2.3^(n-1)).(1+ 2.3^n) ]
= 1/( 1+2.3^(n-1)) - 1/(1+ 2.3^n)
c1+c2+...+cn
= 1/3 - 1/(1+ 2.3^n)