1/(2×3×4)=(1/2)[1/(2×3)－1/(3×4)]
1/(3×4×5)=(1/2)[1/(3×4)－1/(4×5)]
……
1/(8×9×10)=(1/2)[1/(8×9)－1/(9×10)]
则原式=(1/2)[1/(1×2)－1/(9×10)]=(1/2)[(1/2)－1/(90)]=11/45
1\/1x2x3+1\/2x3x4+1\/3x4x5+...+1\/8x9x10如何计算方法
则原式=(1\/2)[1\/(1×2)－1\/(9×10)]=(1\/2)[(1\/2)－1\/(90)]=11\/45
1\/1x2x3+1\/2x3x4+1\/3x4x5+...+1\/8x9x10 求简算
1\/1x2x3+1\/2x3x4+1\/3x4x5+...+1\/8x9x10 =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\/2-1\/90]=1\/2×44\/90 =11\/45
1\/1x2x3+1\/2x3x4+1\/3x4x5+1\/4x5x6+.+1\/8x9x10 等于多少?急!
请1\/1x2x3+1\/2x3x4+1\/3x4x5+1\/4x5x6= =1\/2×(1\/1×2-1\/2×3+1\/2×3-1\/3×4+1\/3×4-1\/4×5+1\/4×5-1\/5×6) =1\/2×(1\/1×2-1\/5×6) =7\/30 1\/1x2x3+1\/2x3x4+1\/3x4x5+...+1\/9x10x11= 1\/n(n+1)(n+2)=1\/2*[1\/n-2\/(n+1)+1\/(n+...
1\/1x2x3+1\/2x3x4+1\/3x4x5+1\/4x5x6+...+1\/8x9x10 等于多少?急急急!!!
1\/1x2x3+1\/2x3x4+1\/3x4x5+1\/4x5x6+...+1\/8x9x10 =1-1\/2-1\/2(1-1\/3)+1\/2-1\/3-1\/2(1\/2-1\/4)+1\/3-1\/4-1\/2(1\/3-1\/5)+1\/4-1\/5-1\/2(1\/4-1\/6)+...+1\/8-1\/9-1\/2(1\/8-1\/10)=1-1\/9-1\/2(1+1\/2-1\/9-1\/10)=173\/360 ...
1\/1X2X3+1\/2X3X4+1\/3X4X5+...+1\/98X99X100(要过程)
1\/1x2x3+1\/2x3x4+1\/3x4x5+--+1\/98x99x100 =(1\/2)*(1\/1*2-1\/2*3)+(1\/2)*(1\/2*3-1\/3*4)+...+(1\/2)(1\/98*99-1\/99*100)=(1\/2)*(1\/1*2-1\/2*2+1\/2*3-1\/3*4+...+1\/98*99-1\/99*100)=(1\/2)*(1\/2-1\/9900)=(1\/2)*(4949\/9900)=4949\/19800.
1\/丨1x2ⅹ3+1\/2x3x4+……+1\/8ⅹ9x10=
1\/1x2x3+1\/2x3x4+1\/3x4x5+1\/4x5x6+...+1\/8x9x10 =1-1\/2-1\/2(1-1\/3)+1\/2-1\/3-1\/2(1\/2-1\/4)+1\/3-1\/4-1\/2(1\/3-1\/5)+1\/4-1\/5-1\/2(1\/4-1\/6)+.+1\/8-1\/9-1\/2(1\/8-1\/10)=1-1\/9-1\/2(1+1\/2-1\/9-1\/10)=173\/360 ...
1\/(1x2x3)+1\/(2x3x4)+1\/(3x4x5)+.
将问题中的计算数字化如下:1\/(1x2x3)+ 1\/(2x3x4) +1\/(3x4x5) +...+ x1\/(98x99x100)由上述式子可以看出，第n项是1\/[n(n+1)(n+2)]，由1\/[n(n+1)(n+2)]与1\/n,1\/(n+1),1\/(n+2)的关系，可以知道下式成立：1\/[n(n+1)(n+2)]=1\/2x[1\/n+1\/(n+2)]-1\/(n...
1\/1x2x3+1\/2x3x4+...+1\/98x99x100 .
1\/1x2x3+1\/2x3x4+...+1\/98x99x100 =1\/2 × [(1\/1x2 - 1\/2x3)+(1\/2x3 - 1\/3x4)+(1\/3x4 - 1\/4x5)+...+(1\/98x99 - 1\/99x100)]=1\/2 × [1\/1x2 - 1\/99x100]=1\/2 × 4994\/9900 =2497\/9900
1\/1x2x3+1\/2x3x4+1\/3x4x5+...+1\/9x10x11=
1\/n(n+1)(n+2)=1\/2*[1\/n-2\/(n+1)+1\/(n+2)]原式=1\/2*(1-2*1\/2+1\/3+1\/2-2*1\/3+1\/4+...+1\/9-2*1\/10+1\/11)=1\/2*(1-1\/2-1\/10+1\/11)=27\/110
1\/1x2x3x4+1\/2x3x4x5+1\/3x4x5x6+1\/4x5x6x7+1\/5x6x7x8+1\/6x7x8x9+1\/...
1\/(1×2×3×4)+1\/(2×3×4×5)+1\/(3×4×5×6)+...+1\/(7×8×9×10)=(1\/3)×［1\/(1×2×3)-1\/(2×3×4)］+(1\/3)×［1\/(2×3×4)-1\/(3×4×5)］+(1\/3)×［1\/(3×4×5)-1\/(4×5×6)］+...+(1\/3)×［1\/(7×8×9)-1\/(8×9×10)］=(...
