51问答网 > 求和：Sn=1*2*3+2*3*4+……+n(n+1)(n+2)

求和：Sn=123+234+……+n(n+1)(n+2)

2024-12-26 08:38:50

推荐回答（2个）

回答1：

an=n(n+1)(n+2)
=((n+3)-(n-1))/4*n(n+1)(n+2)
=[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]/4
sn所有的项都如此裂项分解，加起来，可以狂消
Sn=1*2*3+2*3*4+3*4*5+...+n*(n+1)*(n+2)
=1/4{1*2*3*(4-0)+2*3*4*(5-1)+3*4*5*(6-2)...+n*(n+1)*(n+2)[n+3-(n-1)]}
=[n(n+1)(n+2)(n+3)-0]/4
=n(n+1)(n+2)(n+3)/4

回答2：

Sn=1*2*3+2*3*4+……+n(n+1)(n+2)
=(1³+2³+3³+----+n³)+3(1²+2²+3²+---+n²)+2(1+2+3+----+n)
=n²(n+1)²/4+n(n+1)(2n+1)/2+n(n+1)
=n(n+1)[n(n+1)/4+(2n+1)/2+1)]
=n(n+1)(n+2)(n+3)/4