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Question Number 126097 by amns last updated on 17/Dec/20

Commented by amns last updated on 17/Dec/20

I need someone′s help to solve it!!!

$$\mathrm{I}\:\mathrm{need}\:\mathrm{someone}'\mathrm{s}\:\mathrm{help}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}!!! \\ $$

Answered by Dwaipayan Shikari last updated on 17/Dec/20

(1/2)−(1/(12))−(1/(20))−(1/(30))−(1/(42))−..(1/(210))=1−((1/2)+(1/6)+(1/(12))+(1/(20))+...)+(1/6) =1−Σ_(n=1) ^(14) (1/n)−(1/(n+1))+(1/6)=1−(1−(1/(15)))+(1/6)=(7/(30)) Note: (1/2)+(1/6)+(1/(12))+(1/(20))+...+(1/(210)) =1−(1/2)+(1/2)−(1/3)+(1/3)−(1/4)+(1/4)−(1/5)+...+(1/(14))−(1/(15)) =1−(1/(15))=((14)/(15))

$$\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{20}}−\frac{\mathrm{1}}{\mathrm{30}}−\frac{\mathrm{1}}{\mathrm{42}}−..\frac{\mathrm{1}}{\mathrm{210}}=\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{20}}+...\right)+\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$=\mathrm{1}−\underset{{n}=\mathrm{1}} {\overset{\mathrm{14}} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{6}}=\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{15}}\right)+\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{7}}{\mathrm{30}} \\ $$$$\boldsymbol{\mathcal{N}{ote}}:\:\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{20}}+...+\frac{\mathrm{1}}{\mathrm{210}} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{5}}+...+\frac{\mathrm{1}}{\mathrm{14}}−\frac{\mathrm{1}}{\mathrm{15}} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{15}}=\frac{\mathrm{14}}{\mathrm{15}} \\ $$

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