1*2*3+2*3*4+……+n(n+1)(n+2)=

n(n+1)(n+2)=n^3+3n^2+2n

1^3+2^3+……+n^3=[n(n+1)/2]^2=n^2(n+1)^2/4

1^2+2^2+……+n^2=n(n+1)(2n+1)/6

1+2+……+n=n(n+1)/2

所以1*2*3+2*3*4+……+n(n+1)(n+2)
=n^2(n+1)^2/4+3*n(n+1)(2n+1)/6+2*n(n+1)/2
=n^2(n+1)^2/4+n(n+1)(2n+1)/2+n(n+1)
=[n(n+1)/4][n(n+1)+2(2n+1)+4]
=[n(n+1)/4](n^2+n+4n+2+4)
=[n(n+1)/4](n^2+5n+6)
=n(n+1)(n+2)(n+3)/4

设Sn=1*2*3+2*3*4+……+n(n+1)(n+2),
那么
4Sn=1*2*3*4+2*3*4*4+……+4n(n+1)(n+2)
=2*3*4*(1+4)+3*4*5*4+……+4n(n+1)(n+2)
=2*3*4*5+3*4*5*4+……+4n(n+1)(n+2)
=3*4*5*(2+4)+4*5*6*4+……+4n(n+1)(n+2)
=3*4*5*6+5*6*7*4+……+4n(n+1)(n+2)
……
=n(n+1)(n+2)(n+3)
那么Sn=n(n+1)(n+2)(n+3)/4

1×2×3+2×3×4+…+n（n+1）（n+2） ==(1*2*3*4-0*1*2*3)/4+(2*3*4*5-1*2*3*4)/4+...+(n（n+1）（n+2）*(n+3)-(n-1)n（n+1）（n+2）)/4 =(n（n+1）（n+2）(n+3)-0*1*2*3)/4 =n（n+1）（n+2）（n+3）/4