（1）因为f'(x)=6x-2从而f（x）=3x^2-2x+c
又y=f(x)的图像经过坐标原点,从而c=0,即f（x）=3x^2-2x
所以Sn=3n^2-2n,an=Sn-Sn-1=6n-5
（2）bn=2/[（6n-5）*(6n+1)]=1/3[1/(6n-5)-1/(6n+1)]=1/3[1/an-1/a(n+1)]
从而Tn=b1+b2+.+bn=1/3[1/a1-1/a（n+1）]=2n/(6n+1)