2.由第1小题可得：an=3n-1,an(n+1)=3n+2
那么：bn=1/[an*a(n+1)]
=1/[(3n-1)(3n+2)]
=(1/3)*[1/(3n-1) - 1/(3n+2)]
所以：
数列{bn}的前n项和
Tn=b1+b2+b3+...+bn
=(1/3)*[(1/2 - 1/5) + (1/5 - 1/8) + (1/8 - 1/11) + ...+ 1/(3n-1) - 1/(3n+2)]
=(1/3)*[1/2 - 1/(3n+2)]
=(1/3)*(3n+2-2)/[2(3n+2)]
=n/[2(3n+2)]