题目内容
数列1×4,2×5,3×6,…,n×(n+3),…则它的前n项和Sn=______.
∵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:
.
∴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 |
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