作业帮计算1/x(x+2)+1/(x+2)(x+4)+…+1/(x+98)(x+100)
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/05 01:05:14
计算1/x(x+2)+1/(x+2)(x+4)+…+1/(x+98)(x+100)
计算1/x(x+2)+1/(x+2)(x+4)+…+1/(x+98)(x+100)
计算1/x(x+2)+1/(x+2)(x+4)+…+1/(x+98)(x+100)
1/x(x+2)+1/(x+2)*(x++4)+1/(x+4)*(x+6)+.+1/(x+98)*(x+100)
=[2/x(x+2)+2/(x+2)*(x++4)+2/(x+4)*(x+6)+.+2/(x+98)*(x+100)]/2
=[1/x-1/(x+2)+1/(x+2)-1/(x+4)+……+1/(x+98)-1/(x+100)]/2
=[1/x-1/(x+100)]/2
=(x+100-x)/2x(x+100)
=50/x(x+100)
=50/2*102
=25/102
1/x(x+2)+1/(x+2)(x+4)+…+1/(x+98)(x+100)=1/2（1/x-1/（x+2）+1/（x+2）……-1/（x+100））=50/x（x+100）
此法叫裂项
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