已知数列{an}的各项均为正数,其前n项和为Sn,且an=2√Sn -1,n属于N*,数列b1,b

1.an=2√Sn -1,
n=1时a1=2√a1-1,
(√a1-1)^2=0,a1=1=S1.
n>1时Sn-S=2√Sn-1,
(√Sn-1)^2=S,数列{an}的各项均为正数,
∴√Sn-1=√S,
∴√Sn-√S=1,
∴√Sn=n,Sn=n^2,
an=2n-1,数列{an}是等差数列,不是等比数列.
2.cn=(2n-1)/2^(n-1),
∴Tn=1+3/2+5/2^2+……+(2n-1)/2^(n-1),①
Tn/2=.1/2+3/2^2+……+（2n-3)/2^(n-1)+(2n-1)/2^n,②
①-②,Tn/2=1+2[1/2+1/2^2+……+1/2^（n-1)]-(2n-1)/2^n
=1+2[1-1/2^(n-1)]-(2n-1)/2^n
=3-(2n+3)/2^n,
∴Tn=6-(2n+3)/2^(n-1).