等比数列{an}的前n项和为Sn,已知对任意的n∈N*,点(n,Sn)均为函数y=b^x+r(b>0且bv1≠1,b,r

(1)∵等比数列{a[n]}的前n项和为S[n],对任意的n∈N*,点(n,Sn)均为函数y=b^x+r(b>0且b≠1,b,r均为常数)的图象上
∴S[n]=b^n+r
∵S[n+1]=b^(n+1)+r
∴将上面两式相减,得：a[n+1]=(b-1)b^n
∵a[n]=(b-1)b^(n-1)
∴a[n+1]/a[n]=b
∵a[1]=S[1]=b+r
a[2]=b(b-1)
∴a[2]/a[1]=b(b-1)/(b+r)=b
解得：r=-1
(2)∵当b=2,时记,b[n]=n+1/(4a[n])(n∈N*)
∴b[n]=n+1/[4*2^(n-1)]=n+1/2^(n+1)
∴{b[n]}的前n项和T[n]
=[1+1/2^2]+[2+1/2^3]+[3+1/2^4]+...+[n+1/2^(n+1)]
=(1+2+...+n)+[1/2^2+1/2^3+1/2^4+...+1/2^(n+1)]
=n(n+1)/2+(1/4)[1-1/2^n]/(1-1/2)
=n(n+1)/2+1/2-1/2^(n+1)

高一数列问题