{an}为等差数列,an>0
S5=5a1+10d=35
==> a1+2d=7
即a3=7
a3-1是a1+1和a4的等比中项
(a3-1)^2=(a1+1)(a1+3d)
==>
(7-1)^2=(a1+1)(7+d)
(8-2d)(7+d)=36
-2d^2-6d+20=0
d^2+3d-10=0
解得d=-5或d=2
若d=-5,那么与an恒正矛盾
∴d=2,a1=3
所以an=2n+1bn=an2-3/Sn-n,求数列bn的前n项和TnSn=(3+2n+1)n/2=n^2+2n

bn=[(an)^2-3]/(Sn-n)

=(4n^2+4n-2)/(n^2+n)
=4-2/[n(n+1)]
=4-2[1/n-1/(n+1)]
Tn=b1+b2+...+bn
=4n-2[1-1/2+1/2-1/3+...+1.n-1/(n+1)]
=4n-2[1-1/(n+1)]
=4n-2n/(n+1)