解题思路: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)=
n(n+1)(2n+1)
6
+
3n(n+1)
2
;化简可得答案.
∵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)
=
n(n+1)(2n+1)
6+
3n(n+1)
2
=
n(n+1)(n+5)
3.
答案:
n(n+1)(n+5)
3.
点评:
本题考点: 数列的求和.
考点点评: 本题考查数列的性质和应用,解题时要认真审题,仔细求解.