Question

初中数学题(1*2*3+2*3*4+3*4*5+……n*(n+1)(n+2))=?

Answer 1

1*2*3=1/4(1*2*3*(4-0)
2*3*4=1/4(2*3*4*(5-1)
......
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)]}
=1/4{1*2*3*4+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)}原式= n*(n+1)*(n+2）*（n+3）/4
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Answer 2

对其通项进行分析n(n+1)(n+2)=1/4[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]
故有
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)]
这n项左边加起来就是要求的问题
右边加起来就结果，观察可知到右边只剩下1/4[n(n+1)(n+2)(n+3)
故Sn=1/4[n(n+1)(n+2)(n+3)

Answer 3

用整数裂项法,把每个n×(n+1)×(n+2)写成1/4[n×(n+1)×(n+2)×(n+3)-(n-1)×n×(n+1)×(n+2)]
原式=1/4[1×2×3×4-0×1×2×3+2×3×4×5-1×2×3×4-n×(n+1)×(n+2)×(n+3)-(n-1)×n×(n+1)×(n+2)]
=1/4[n×(n+1)×(n+2)×(n+3)]

Answer 4

1.Sn=1/4[n(n+1)(n+2)(n+3)
2.1/4[n×(n+1)×(n+2)×(n+3)-(n-1)×n×(n+1)×(n+2)]
原式=1/4[1×2×3×4-0×1×2×3+2×3×4×5-1×2×3×4-n×(n+1)×(n+2)×(n+3)-(n-1)×n×(n+1)×(n+2)]
=1/4[n×(n+1)×(n+2)×(n+3)]

Answer 5

原式= n*(n+1)*(n+2）*（n+3）/4