已知数列{an}的前几项和为Sn,点（n,Sn）（n∈N*）均在二次函数f（x）=3x²－2x的图像上.⑴求数列｛an

(1) Sn=3n²-2n S(n-1)=3(n-1)²-2(n-1) an=Sn-S(n-1)=3n²-2n-3(n-1)²+2(n-1)=6n-5
(2) bn=3/[（6n-5）（6n+1）]=1/2×[1/(6n-5)-1/(6n+1)] Tn=1/2[1/1-1/7+1/7-1/13+…-1/(6n+1)]=3n/(6n+1)第二问是:设bn=3/an·a﹙n﹢1﹚,求数列﹛bn﹜的前n项和Tn.bn=3/[(6n-5)(6n+1)]=1/2×[1/(6n-5)-1/(6n+1)]Tn=1/2[1/1-1/7+1/7-1/13+…-1/(6n+1)]=3n/(6n+1)