A．3n2+5n(n+1)(n+2)B．2n2+5n(n+1)(n+3)C．3n2+2n(n+2)(n+3)D．2n2+

an＝
1+2+…+n
n+1=

n(n+1)
2
n+1=[n/2]．
∴bn＝
1

n
2•
n+2
2=2(
1
n−
1
n+2)．
∴S_n=2[(1−
1
3)+(
1
2−
1
4)+(
1
3−
1
5)+…+(
1
n−1−
1
n+1)+(
1
n−
1
n+2)]
=2(1+
1
2−
1
n+1−
1
n+2)
=
3n2+5n
(n+1)(n+2)．
故选A．

点评：
本题考点： 数列的求和；等差数列的前n项和．

考点点评： 熟练掌握等差数列的前n项和公式、“裂项求和”等是解题的关键．