已知数列{an}中，Sn=an/2+1/{(an)-1}[n>=1](n€N*),求数列的{an}的通项公式？帮帮忙！

an=2a(n-1)+2^n-1
an/2^n=a(n-1)/2^(n-1) +1-1/2^n
an/2^n-a(n-1)/2^(n-1)=1-1/2^n

a2/2^2-a1/2=1-1/2^2
a3/2^3-a2/2^2=1-1/2^3
...
an/2^n-a(n-1)/2^(n-1)=1-1/2^n

左右两边分别相加：
an/2^n-a1/2=(n-1)-[1/2^2+..+1/2^n]
an/2^n=a1/2+(n-1)-1/2^2*[1-1/2^(n-1)]/(1-1/2)
=a1/2+(n-1)-(1/2-1/2^n)
=a1/2+n-3/2+1/2^n
a1=5 (a4求a3-a2-a1,过程略）
an=(n+1)*2^n+1

Sn=a1+a2+..+an=n+2^1+2^2+..+2^n+1*2^1+2*2^2+..+n*2^n
=n+2*(1-2^n)/(1-2)+1*2^1+2*2^2+..+n*2^n

bn=1*2^1+2*2^2+..+n*2^n
2bn=1*2^2+........+(n-1)*2^n+n*2^(n+1)
相减：
-bn=2+2^2+........+2^n-n*2^(n+1)
=2*(1-2^n)/(1-2)-n*2^(n+1)
bn=-2*(1-2^n)/(1-2)+n*2^(n+1)

Sn=n+n*2^(n+1)

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