已知数列{an}的前n项和Sn=n(n+1)/2 (n∈N*) (1)求数列{an}的通项公式 (2)设数列{bn}满足(2an-1)(2^(bn)-1)=1,Tn为{bn}的前n项和,求证：

(1) an=Sn-S(n-1)n(n+1)/2-n(n-1)/2=n
(2n-1)(2^bn-1)=1
=>2^bn-1=1/(2n-1)
2^bn=2n/(2n-1)
bn=log(2)[1+1/(2n-1)]
Tn=log(2)[(1+1/1)(1+1/3)...(1+1/(2n-1)]
欲证2Tn>log2(2an+1),只需证明
2log(2)[(1+1/1)(1+1/3)...(1+1/(2n-1)]>log(2)(2n+1)
<=>[(1+1/1)(1+1/3)...(1+1/(2n-1)]^2>2n+1
上式可以用数学归纳法证明.
n=1 (1+1)^2>2+1=3,显然成立.
假设n=n时成立.
[(1+1/1)(1+1/3)...(1+1/(2n-1)]^2>2n+1
当n=n+1时,
左边=[(1+1/1)(1+1/3)...(1+1/(2n-1)]^2[1+1/(2n+1)]^2>(2n+1)[1+1/(2n+1)]^2=(2n+2)^2/(2n+1)=2n+3+1/(2n+1)>2n+3
显然成立.
反推回去,本题得证.