1.不等式f(x)1时,Sn-1=3（n-1）^2-2(n-1)
得到an=6n-5 a1也符合该式
所以an=6n-5
2.bn=3/anan+1=1/2[1/（6n-5）-1/（6n+1）]
故Tn=1/2(1-17+1/7+^-1/(6n+1))=1/2[1-1/(6n+1)]=3n/(6n+1)