作业帮1/(5×9)+1/(9×13)+1/(13×17)+⋯+1/(101×105)
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/14 21:52:58
1/(5×9)+1/(9×13)+1/(13×17)+⋯+1/(101×105)
1/(5×9)+1/(9×13)+1/(13×17)+⋯+1/(101×105)
1/(5×9)+1/(9×13)+1/(13×17)+⋯+1/(101×105)
原式=1/4*(1/5-1/9+1/9-1/13+1/13-1/17+……+1/101-1/105）
可以发现,中间的全部可以抵消.
=1/4*（1/5-1/105）
=1/4*4/21
=1/21
1/(5×9)+1/(9×13)+1/(13×17)+⋯+1/(101×105)
=(1/4)*(1/5-1/9+1/9-1/13+....+1/101-1/105)
=(1/4)*(1/5-1/105)
=(1/4)*(20/105)
=1/21
上边是对的
首先观察一下，分母数字刚好是4为公差的等差数列，
1/5-1/9=(9-5)/(5×9)=4/(5×9)
1/9-1/13=(13-9)/(9×13)=4/(9×13)
...
1/101-1/105=(105-101)/(101×105)=4/(101×105)
所以：
原式=1/(5×9)+1/(9×13)+1/(13×17)+...+1/(10...
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首先观察一下，分母数字刚好是4为公差的等差数列，
1/5-1/9=(9-5)/(5×9)=4/(5×9)
1/9-1/13=(13-9)/(9×13)=4/(9×13)
...
1/101-1/105=(105-101)/(101×105)=4/(101×105)
所以：
原式=1/(5×9)+1/(9×13)+1/(13×17)+...+1/(101×105)
=(1/4)*[4/(5×9)+4/(9×13)+4/(13×17)+...+4/(101×105)]
=(1/4)*(1/5-1/9+1/9-1/13+1/13-1/17+...+1/101-1/105)
=(1/4)*(1/5-1/105)
=(1/4)*(4/21)
=1/21
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