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Question-119032




Question Number 119032 by Algoritm last updated on 21/Oct/20
Answered by 1549442205PVT last updated on 22/Oct/20
1−(1/2)+(1/3)−(1/4)+...−(1/(2n))=(1/(n+1))+...+(1/(2n))  Put S=1−(1/2)+(1/3)−(1/4)+...−(1/(2n))  ⇒ S=1+(1/2)+(1/3)+(1/4)+...+(1/(2n−1))+(1/(2n))  −2((1/2)+(1/4)+...+(1/(2n)))  =1+(1/2)+(1/3)+(1/4)+...+(1/(2n−1))+(1/(2n))  −(1+(1/2)+(1/3)+...+ (1/n))=(1/(n+1))+...+(1/(2n))  (Q.E.D)
$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+…−\frac{\mathrm{1}}{\mathrm{2n}}=\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}+…+\frac{\mathrm{1}}{\mathrm{2n}} \\ $$$$\mathrm{Put}\:\mathrm{S}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+…−\frac{\mathrm{1}}{\mathrm{2n}} \\ $$$$\Rightarrow\:\mathrm{S}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{\mathrm{2n}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2n}} \\ $$$$−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{\mathrm{2n}}\right) \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{\mathrm{2n}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2n}} \\ $$$$−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\:\frac{\mathrm{1}}{\mathrm{n}}\right)=\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}+…+\frac{\mathrm{1}}{\mathrm{2n}} \\ $$$$\left(\boldsymbol{\mathrm{Q}}.\boldsymbol{\mathrm{E}}.\boldsymbol{\mathrm{D}}\right) \\ $$

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