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Question-157781




Question Number 157781 by amin96 last updated on 27/Oct/21
Answered by mr W last updated on 27/Oct/21
100A=((1/1)−(1/(101)))+((1/2)−(1/(102)))+...+((1/(25))−(1/(125)))  25B=((1/1)−(1/(26)))+((1/2)−(1/(27)))+...+((1/(100))−(1/(125)))  25B−100A=(1+(1/2)+(1/3)+...+(1/(125)))−(1+(1/2)+(1/3)+...+(1/(125)))=0  25B=100A  (A/B)=((25)/(100))=(1/4)
$$\mathrm{100}{A}=\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{101}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{102}}\right)+…+\left(\frac{\mathrm{1}}{\mathrm{25}}−\frac{\mathrm{1}}{\mathrm{125}}\right) \\ $$$$\mathrm{25}{B}=\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{26}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{27}}\right)+…+\left(\frac{\mathrm{1}}{\mathrm{100}}−\frac{\mathrm{1}}{\mathrm{125}}\right) \\ $$$$\mathrm{25}{B}−\mathrm{100}{A}=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{\mathrm{125}}\right)−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{\mathrm{125}}\right)=\mathrm{0} \\ $$$$\mathrm{25}{B}=\mathrm{100}{A} \\ $$$$\frac{{A}}{{B}}=\frac{\mathrm{25}}{\mathrm{100}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

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