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

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...

原式=（1/2）(1/(3*1)) + .+(1/10)(1/(9*11))= (1/2)(1/2)(1/1 -1/3)+.+(1/2)(1/10)(1/9 -1/11)所以原式*2 = (1/2)(1/1)- (1/2)(1/3).+(1/10)(1/9)-(1/10)(1/11)= [(1/2)(1/1)+.+(1/10)(1/9)]-[(1/2)(1/3)+(1/10)(1...

原式=(1/1)*[1/（2*3）]+(1/2)*[1/()3*4]+.+(1/9)[1/(10*11)]=(1/1)*(1/2-1/3)+(1/2)*(1/3-1/4)+.+(1/9)(1/10-1/11)=1/(1*2)-1/(1*3)+1/(2*3)-1/(2*4)+.+1/(9*10)-1/(9*11)=[1/(1*2)+1/(2*3)+.+1/(9*10)]-[1/(1*...

原式=(1/1)*[1/（2*3）]+(1/2)*[1/()3*4]+...+(1/9)[1/(10*11)]=(1/1)*(1/2-1/3)+(1/2)*(1/3-1/4)+...+(1/9)(1/10-1/11)=1/(1*2)-1/(1*3)+1/(2*3)-1/(2*4)+...+1/(9*10)-1/(9*11)=[1/(1*2)+1/(2*3)+...+1/(9*10)...

n+1)-1/(n+1)*(n+2))1/(n*(n+1)=1/n-1/(n+1)代入 (1/2)*(1/n-2/(n+1)+1/(n+2))这样的话差不多好了，叠加，消去，就可以得到结果 1-1/2-1/100+1/101=5049/10100 再乘1/2 就是5049/20200 楼上打得比我快,汗了 看到楼下的解法，我发现，我干了蠢事 ...

1/(2*3*4)=1/2*[1/2*3-1/3*4]=1/2*[(1/2-1/3)-(1/3-1/4)]=1/2*[1/2-2/3+1/4]1/(3*4*5)=1/2*[1/3*4-1/4*5]=1/2*[(1/3-1/4)-(1/4-1/5)]=1/2*[1/3-2/4+1/5]...1/(26*27*28)=1/2*[1/26*27-1/27*28]=1/2*[(1/...

解：1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+……+1/(98*99*100)=(1/2)*(4-3)/(3*4)+(1/3)*(5-4)/(4*5)+(1/4)*(6-5)/(5*6)+……+(1/98)*(100-99)*(99*100)=(1/2)*(1/3-1/4)+(1/3)*(1/4-1/5)+(1/4)*(1/5-1/6)+……+(1/98)*(1/...

1*2*3分之一+2*3*4分之一+3*4*5分之一...+8*9*10分之一 =1/2*（1/1*2-1/2*3+1/2*3-1/3*4+1/3*4-1/4*5+...+1/8*9-1/9*10）=1/2*（1/1*2-1/9*10）=1/2*（1/2-1/90）=1/2*22/45 =11/45 ...

运用公因数，进行通分，然后相加，得到结果。

1/(2×3×4)=1/2[1/(2×3)-1/(3×4)]...1/(20×21×22)=1/2[1/(20×21)-1/(21×22)]左右两边分别相加，中间项都是一正一负消去了，只剩首尾项 1/2[1/(1×2)-1/(21×22)]=1/4[1-1/21×11]=115/462 即1/(1×2×3)+1/(2×3×4)+···+1/(20×21×...