Sn=123+234+345+...+n(n+1)(n+2)求Sn

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Sn=1*2*3+2*3*4+3*4*5+...+n(n+1)(n+2)求Sn
Sn=1*2*3+2*3*4+3*4*5+...+n(n+1)(n+2)求Sn

Sn=1*2*3+2*3*4+3*4*5+...+n(n+1)(n+2)求Sn
n*(n+1)*(n+2)=1/4*n*(n+1)*(n+2)[n+3-(n-1)]
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)]}
如此裂项相消
原式= n*(n+1)*(n+2）*（n+3）/4

可以用积分算，你可以对x^3+x^4+……+x^（n+3），求和，然后把和函数求三次导，最后令x=1。

an=n(n+1)(n+2)=n^3+3n^2+2n
Sn=1^3+2^3+..n^3+3(1^2+2^2+...n^2)+2(1+2+...+n)
因为1^3+2^3+3^3+4^3+……+n^3=[n(n+1)/2]^2
1^2+2^2+3^2+4^4+……+n^2=
1+2+3+4+……+n=n(n+1)/2
Sn=[n(n+1)/2]^2+n(n+1)(2n+1)/2+n(n+1)

Sn’=1*2*3*4+2*3*4*5+...+(n-1)n(n+1)(n+2)+n(n+1)(n+2)(n+3) (1)
Sn'=1*2*3*4+...+(n-2)(n-1)n(n+1)+(n-1)n(n+1)(n+2)+n(n+1)(n+2)(n+3) (2)
(1)-(2)得：0=1*2*3*4+4*2*3*4+4*3*4*5+...+...

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Sn’=1*2*3*4+2*3*4*5+...+(n-1)n(n+1)(n+2)+n(n+1)(n+2)(n+3) (1)
Sn'=1*2*3*4+...+(n-2)(n-1)n(n+1)+(n-1)n(n+1)(n+2)+n(n+1)(n+2)(n+3) (2)
(1)-(2)得：0=1*2*3*4+4*2*3*4+4*3*4*5+...+4(n-1)n(n+1)+4n(n+1)(n+2)-n(n+1)(n+2)(n+3)
0=4*{1*2*3+2*3*4+3*4*5+...+(n-1)n(n+1)+n(n+1)(n+2)}-n(n+1)(n+2)(n+3)
4*Sn=n(n+1)(n+2)(n+3)
所以Sn=n(n+1)(n+2)(n+3)/4

收起

Sn=sum(n*(n+1)*(n+2))=sum(n^3)+sum(3n^2)+sum(2n)
=[n(n+1)/2]^2+n(n+1)(2n+1)/2+n(n+1)

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