求和sn＝1×2×3+2×3×4+……+n(n+1)(n+2)评给五星
问题描述：
求和sn＝1×2×3+2×3×4+……+n(n+1)(n+2)
评给五星
答
因为n(n+1)(n+2)=n^3+3n^2+2n
因为：1^3+2^3+........=n^3=[n(n+1)/2]^2
1^2+2^2+......+.n^2=n（n+1)(2n+1)/6
1+2+3+......+n=n(n+1)/2
所以原式等于sn＝1×2×3+2×3×4+……+n(n+1)(n+2)=[n(n+1)/2]^2+n（n+1)(2n+1)/2
+n(n+1) =n(n+1)(n+2)(n+3)/4
答
n(n+1)(n+2)= [-(n-1)n(n+1)(n+2)+n(n+1)(n+2)(n+3)]/4
代入原式＝n(n+1)(n+2)(n+3)/4
答
利用组合数公式
n(n+1)(n+2)=6*C(n+2,3)
Sn=6[C(3,3)+C(4,3)+C(5,3)+.+C(n+2,3)]
Sn=6[C(4,4)+C(4,3)+C(5,3)+.+C(n+2,3)]
连续 利用公式C(n,m)+C(n,m-1)=C(n+1,m)
Sn=6[C(5,4)+C(5,3)+.+C(n+2,3)]
=6[C(6,4)+.+C(n+2,3)]
=6C(n+3,4)
=(n+3)(n+2)(n+1)n*6/24
=(n+3)(n+2)(n+1)n/4