Question

求数列{n(n+1)(2n+1)}的前n项和

Answer 1

1)
1^2
+
2^2
+
3^2
+
……
+
n^2
=
n(n+1)(2n+1)/6
2)
1^3
+
2^3
+3^3
+
……+
n^3
=
[n(n+1)/2]^2
因此可以把所求式子展开,然后利用上面的2个公式
n(n+1)(2n+1)
=
(n^2+n)(2n+1)
=
2n^3
+3n^2
+n
Sn
=
2*(1^3+2^3+……+n^3)
+
3*(1^2+2^2+
……+n^2)
+
(1+2+……+n)
=
2*[n(n+1)/2]^2
+
3*n(n+1)(2n+1)/6
+
n(n+1)/2
=
[n*(n+1)]^2/2
+
n(n+1)(2n+1)/2
+
n(n+1)/2
提出
n(n+1)/2
=
[n(n+1)/2]
*
[n(n+1)
+
(2n+1)
+
1]
=
[n(n+1)/2]
*
(n^2
+3n+2)
=
n
*
(n+1)^2
*
(n+2)
/2