Question

（3）1×2×3+2×3×4+…n(n+1)(n+2)怎么解

Answer 1

整数裂项
1×2×3=(1×2×3×4-0×1×2×3)÷4
2×3×4=(2×3×4×5-1×2×3×4)÷4
......
n(n+1)(n+2)=[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]÷4
中间各项消去,
1×2×3+2×3×4+…n(n+1)(n+2)=n(n+1)(n+2)(n+3)÷4

追问

后面是n-z*n*n+1

追答

神马意思?

追问

（3）1×2×3+2×3×4+…（n-1）*n*(n+1）这是原式
答案我已经算出来了。谢谢！刚算出来
麻烦你了。
888888888886，后会无期！

追答

井

Answer 2

n(n+1)(n+2)
=n(n²+3n+2)
=n³+3n²+n

∴1×2×3+2×3×4+…n(n+1)(n+2)
=(1³+3*1²+1)+(2³+3*2²+2)+……+(n³+3n²+n)
=(1³+2³+……+n³)+3(1²+2²+……n²)+(1+2+……n)
=[n(n+1)/2]^2+3[n(n+1)(2n+1)/6]+n(n+1)/2
=n²(n+1)²/4+n(n+1)(2n+1)/2+n(n+1)/2
=n(n+1)[n(n+1)+2(2n+1)+2]/4
=n(n+1)(n²+n+4n+2+2)/4
=n(n+1)(n²+5n+4)/4
=n(n+1)(n+1)(n+4)/4
=n(n+1)²(n+4)/4

追问

后面是n-z*n*n+1