已知各项均为正数的数列{an}满足a1=3,且(2an+1-an)/(2an-an+1)=anan+1

解：(1)∵(2a[n+1]-a[n])/(2a[n]-a[n+1])=a[n]a[n+1],n∈N
∴2a[n+1]-a[n]=a[n]a[n+1](2a[n]-a[n+1])
∵{a[n]}是各项均为正数的数列
∴两边同除以a[n]a[n+1]，得：2/a[n]-1/a[n+1]=2a[n]-a[n+1]
即：a[n+1]-1/a[n+1]=2(a[n]-1/a[n])
∵a1=3
∴{a[n]-1/a[n]}是首项为a[1]-1/a[1]=8/3，公比为2的等比数列
即：a[n]-1/a[n]=8*2^(n-1)/3=2^(n+2)/3

∴a[n]^2-a[n]2^(n+2)/3-1=0
a[n]=2^(n+1)/3±√[2^(2n+2)/9+1]
∵[2^(n+1)/3]^2＜[2^(n+1)/3]^2+1=[2^(2n+2)/9+1]
又∵{a[n]}是各项均为正数的数列
∴a[n]=√[2^(2n+2)/9+1]+2^(n+1)/3

(2)∵a[n]=√[2^(2n+2)/9+1]+2^(n+1)/3
设：c[n]=2^(n+1)/3，则：c[n]^2=2^(2n+2)/9
∴a[n]=√[c[n]^2+1]+c[n]
∵有理化1/a[n]，有：1/a[n]=√[c[n]^2+1]-c[n]
∴a[n]^2+(1/a[n])^2=4c[n]^2+2

∵S[n]=a[1]^2+a[2]^2+…+a[n]^2
T[n]=1/a[1]^2+1/a[2]^2+…+1/a[n]^2
∴S[n]+T[n]
=(a[1]^2+1/a[1]^2)+(a[2]^2+1/a[2]^2)+…+(a[n]^2+1/a[n]^2)
=(4c[1]^2+2)+(4c[2]^2+2)+...+(4c[n]^2+2)
=4(c[1]^2+c[2]^2+...c[n]^2)+2n
=4(2^(2*1+2)/9+2^(2*2+2)/9+...+2^(2*n+2)/9)+2n
=4*16[(4^n-1