1/（1*2*3）+1/（2*3*4）+1/（3*4*5）+...+1/（9*10*11）简便运算

Question

请数学好的人帮帮忙，谢谢！！！

Answer 1

1/（1*2*3）+1/（2*3*4）+1/（3*4*5）+...+1/（9*10*11）
=1/2[1/(1×2)-1/(2×3)+1/(2×3)-1/(3×4)+...+1/(9×10)-1/(10×11)]
=1/2×[1/(1×2)-1/(10×11)]
=1/2×(1/2-1/110)
=1/2×54/110
=27/110

公式：1/n(n+1)(n+2)=1/2[1/n(n+1)-1/(n+1)(n+2)]

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Answer 2

∵1/(1*2*3)=(1/2)[(1-1/2)+(1/3-1/2)]
1/(2*3*4)=(1/2)[(1/2-1/3)+(1/4-1/3)]
1/(3*4*5)=(1/2)[(1/3-1/4)+(1/5-1/4)]
.........
.........
1/[n*(n+1)*(n+2)]=(1/2)[(1/n-1/(n+1))+(1/(n+2)-1/(n+1))]
∴1/1*2*3+1/2*3*4+1/3*4*5+...+1/n*(n+1)*(n+2)
=(1/2)[(1-1/(n+1))+(1/(n+2)-1/2)]
=(1/2)[1/2-1/(n+1)+1/(n+2)]
=(1/2){1/2-1/[(n+1)(n+2)]}
=(1/2)n(n+3)/[2(n+1)(n+2)]
=n(n+3)/[4(n+1)(n+2)]
1/（1*2*3）+1/（2*3*4）+1/（3*4*5）+...+1/（9*10*11）
=9*（9+3）/[4(9+1)(9+2)]
=108/440
=27/110

Answer 3

应该是用裂项做的吧！