计算1/[x(x+3)]+1/[(x+3)(x+6)]+......+1/[(x+15)(x+18)]=?

来源：百度知道编辑：UC知道时间：2024/05/10 12:31:42

我需要过程哦!
为什麽开始时是1/3？？？？？？？？？？

1/[x(x+3)]+1/[(x+3)(x+6)]+......+1/[(x+15)(x+18)]
=1/3[1/x-1/(x+3)]+[1/(x+3)-1/(x+6)]+......+[1/(x+15)-1/(x+18)]
=1/3[1/x-1/(x+3)+1/(x+3)-1/(x+6)+......+1/(x+15)-1/(x+18)]
=1/x-1/(x+18)
=1/3*(x+18-x)/x(x+18)
=6/(x^2+18x)

1/[x(x+3)]+1/[(x+3)(x+6)]+......+1/[(x+15)(x+18)]
={3/[x(x+3)]+3/[(x+3)(x+6)]+......+3/[(x+15)(x+18)]}/3
={[(1/x-1/(x+3)]+[1/(x+3)-1/(x+6)]+......+[1/(x+15)-1/(x+18)]}/3
=[1/x-1/(x+18)]/3
=[(x+18-x)/(x(x+18)]/3
=6/(x^2+18x)

用裂项法
原式=1/3*【1/x-1/（x+3）+1/（x+3）-1/（x+6）+...1/（x+15）-1/（x+18）】
中括号里的前后消掉，
原式=1/3*【1/x-1/（x+18）】
=-6/【x（x+18)]

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