已知数列{An}的前n项和为Sn,点(n,Sn) (n属于N+）在函数f(x)=3x^2-2x图象上

Sn=3n^2-2n
Sn-1=3(n-1)^2-2(n-1)
an=6n-5
bn=3/(6n-5)(6n+1)=1/2(1/(6n-5)-1/(6n+1))
tn=1/2(1/a1-1/(6n+1))=1/2(1-1/(6n+1))=3n/(6n+1)
tn-1/2=1/(6n+1)100 n>33/2 n=34