1×2+2×3×4……+n(n+1)(n+2)=

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1×2+2×3×4……+n(n+1)(n+2)=

1×2+2×3×4……+n(n+1)(n+2)=
1×2+2×3×4……+n(n+1)(n+2)=

1×2+2×3×4……+n(n+1)(n+2)=
1×2×3=1/4[1×2×3×4-0×1×2×3]
2×3×4=1/4[2×3×4×5-1×2×3×4]
3×4×5=1/4[3×4×5×6-2×3×4×5]
.
n(n+1)(n+2)=1/4[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]
所以1×2×3+2×3×4+3×4×5+...+n(n+1)(n+2)
=1/4[1×2×3×4-0×1×2×3+2×3×4×5-1×2×3×4+3×4×5×6-2×3×4×5+.+n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]
=1/4n(n+1)(n+2)(n+3)

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