已知数列{an}的各项均为正数,其前n 项和为Sn.且满足2Sn=an^2+an(n∈N*). 1、求数列an的通项公式

n=1时,2S1=2a1=a1²+a1
a1²-a1=0 a1(a1-1)=0
a1=0(各项均为正数,舍去)或a1=1
n≥2时,
2Sn=an²+an
2Sn-1=a(n-1)²+a(n-1)
2Sn-2Sn-1=2an=an²+an-a(n-1)²-a(n-1)
an²-a(n-1)²-an-a(n-1)=0
[an+a(n-1)][an-a(n-1)]-[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-1]=0
数列各项均为正,an+a(n-1)恒>0,要等式成立,只有an-a(n-1)=1,为定值.
数列{an}是以1为首项,1为公差的等差数列.
an=1+n-1=n
数列{an}的通项公式为an=n
bn=n×(1/2)^an=n/2^n
Tn=b1+b2+b3+...+bn=1/2^1+2/2^2+3/2^3+...+n/2^n
Tn/2=1/2^2+2/2^3+...+(n-1)/2^n+n/2^(n+1)
Tn-Tn/2=Tn/2=1/2^1+1/2^2+1/2^3+...+1/2^n -n/2^(n+1)
=(1/2)(1-1/2^n)/(1-1/2) -n/2^(n+1)
=1-1/2^n -n/2^(n+1)
Tn=2 -1/2^(n-1) -n/2^n