已知公差不为0的等差数列{an}，满足a3=9，a1a7的等比中项是a2，,等差数列{an}的前n项和sn，

设等差数列{an}公差为d≠0
∵a3=9，a1a7的等比中项是a2
∴{a1+2d=9
{a1(a1+6d)=(a1+d)²
(9-2d)(9+4d)=(9-d)²
d²-4d=0
∵d≠0
∴d=4,a1=1
∴an=1+4(n-1)=4n-3
Sn=(a1+an)*n/2=2n²-n
(2)
【bn=1/[(an+1)²-4]=1/[(an-1)(an+3)]
=1/[(4n-4)(4n)有问题，n=1无意义】
n+1是项号吗？
bn=1/[a²(n+1)-4]
=1/[(4n+1)²-4]
=1/[(4n-1)(4n+3)]
=1/4[1/(4n-1)-1/(4n+3)]
∴{bn}的前n项和
Tn=1/4[1/3-1/7+1/7-1/11+......+1/(4n-1)-1/(4n+3)]
=1/4[1/3-1/(4n+3)]
=1/12-1/(16n+12)
∵1/(16n+12)>0
∴1/12-1/(16n+12)<1/12
即Tn<12

a3=9，∴a1+2d=9 ①

a1a7的等比中项是a2,∴a²2=﹙a1＋d﹚²=a1·﹙a1＋6d﹚②.

由①②得a1=1.d=4
∴an=4n-3,sn=2n²-n.