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1-4-1-12-1-24-1-2n-n-1-




Question Number 146009 by iloveisrael last updated on 10/Jul/21
 (1/4)+(1/(12))+(1/(24))+...+(1/(2n(n+1)))=?
$$\:\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{24}}+…+\frac{\mathrm{1}}{\mathrm{2n}\left(\mathrm{n}+\mathrm{1}\right)}=? \\ $$
Answered by puissant last updated on 10/Jul/21
=Σ_(k=1) ^n  (1/(2k(k+1)))  =(1/2)Σ_(k=1) ^n ((1/k)−(1/(k+1))) telescopie..  ⇒ S=(1/2)(1−(1/(n+1)))=(1/2)(((n+1−1)/(n+1)))  ⇒S=(n/(2(n+1)))..
$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2k}\left(\mathrm{k}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right)\:\mathrm{telescopie}.. \\ $$$$\Rightarrow\:\mathrm{S}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{n}+\mathrm{1}−\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right) \\ $$$$\Rightarrow\mathrm{S}=\frac{\mathrm{n}}{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)}.. \\ $$

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