求数列{n(n+1)(2n+1)}的前n项和

Question

求数列{n(n+1)(2n+1)}的前n项和

Answer 1

先看上面这个,给出了2个公式及其推导过程.
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