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) + ...+...an(n+1)=3n+2是什么意思an(n+1)？？是（a(n+1)）吧哦哦多了个n哈，a(n+1)=3n+2 即第n+1项！ 抱歉。哦，我看错了懂了谢谢