∵2Sn=3an-2n (1)
∴2S(n+1)=3a(n+1)-2(n+1) (2)
(2)-(1):
2a(n+1)=2S(n+1)-Sn
=3a(n+1)-2(n+1)-3an+2n
=3a(n+1)-3an-2
∴a(n+1)=3an+2
∴[1+a(n+1) ]/(1+an)=3
∴ {1+an}是等比数列,公比为3
∵2a1=2S1=3a1-2
∴a1=2
∴1+an=3^n
∴an=3^n-1
2
∵ an/[1+a(n+1)]=(3^n-1)/[1+3^(n+1)-1]
=(3^n-1)/3^(n+1)=1/3-1/3^(n+1)
∴Tn=a1/(1+a2)+a2/(1+a3)+……+an/(1+a(n+1)),
=(1/3-1/3^2)+(1/3-1/3^3)+.+[1/3-1/3^(n+1)]
=n/3-[1/3^2+1/3^3+.+1/3^(n+1)]
=n/3-1/9(1-/3^n)/(1-1/3)
=n/3-1/6+1/6*1/3^n (∵1/6*1/3^n>0)
>n/3-1/6=(2n-1)/6