数列{an},{bn}满足anbn=1,an=n*n(n的平方)+3n+2,则{bn}的前10项之和为（）

an= n^2+3n+2
=(n+1)(n+2)
bn = 1/[(n+1)(n+2)]
= 1/(n+1) -1/(n+2)
S10
=b1+b2+..+b10
= (1/2-1/3) +(1/3-1/4) +..+(1/11-1/12)
=1/2-1/12
=5/12