Question

已知二次函数f(x)=3x^2-2x,数列an的前n项和为Sn ,

Answer 1

1,Sn=3n^2-2n,Sn-1=3（n-1）^2-2（n-1）,an=Sn-Sn-1=6n-5

2,bn=1/[an*a(n+1)]=1/[(6n-5)(6n+1)]=1/6(1/(6n-5)-1/(6n+1)],

Tn=b1+b2+b3+...=1/6-1/7+1/7-1/13+.+1/6(1/(6n-5)-1/(6n+1)]=1/6-1/[6(6n+1)],