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Question Number 154896 by maged last updated on 22/Sep/21

find the following ?please  S=Σ_(k=1) ^n (1/(k(k+1)(k+2)))

$${find}\:{the}\:{following}\:?{please} \\ $$$${S}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)} \\ $$

Commented by benhamimed last updated on 23/Sep/21

Σ(1/(k(k+1)(k+2)))=Σ(1/2)((1/k)−(2/(k+1))+(1/(k+2)))  =(1/2)(1−(2/2)+(1/3)+(1/2)−(2/3)+(1/4)+(1/3)−(2/4)+(1/5)+...  (1/(n−1))−(2/n)+(1/(n+1))+(1/n)−(2/(n+1))+(1/(n+2)))  =(1/2)((1/2)−(1/(n+1))+(1/(n+2)))

$$\Sigma\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}=\Sigma\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{k}}−\frac{\mathrm{2}}{{k}+\mathrm{1}}+\frac{\mathrm{1}}{{k}+\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{2}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+...\right. \\ $$$$\left.\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{2}}{{n}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}}−\frac{\mathrm{2}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}\right) \\ $$

Commented by maged last updated on 23/Sep/21

Thank you

$${Thank}\:{you} \\ $$

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