
分析:由题设条件知:an=n×(n+3)=n2+3n,Sn=(1+3×1)+(4+3×2)+(9+3×3)+…+(n2+3n)=(12+22+32+…+n2)+3(1+2+3+…+n)=
+
;化简可得答案.解答:∵an=n×(n+3)=n2+3n,
∴Sn=a1+a2+a3+…+an
=(1+3×1)+(4+3×2)+(9+3×3)+…+(n2+3n)
=(12+22+32+…+n2)+3(1+2+3+…+n)
=
+
=
.答案:
.点评:本题考查数列的性质和应用,解题时要认真审题,仔细求解.