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Question Number 170572 by daus last updated on 27/May/22

Answered by mr W last updated on 27/May/22

((k+2)/(k!+(k+1)!+(k+2)!)) =((k+2)/(k!(k^2 +4k+4))) =(1/(k!(k+2))) =((k+1)/((k+2)!)) =((k+2−1)/((k+2)!)) =(1/((k+1)!))−(1/((k+2)!)) Σ_(k=1) ^n =(1/((1+1)!))−(1/((n+2)!))=(1/2)−(1/((n+2)!)) n=999 Σ_(k=1) ^(999) =(1/2)−(1/(1001!)) ✓

$$\frac{{k}+\mathrm{2}}{{k}!+\left({k}+\mathrm{1}\right)!+\left({k}+\mathrm{2}\right)!} \\ $$$$=\frac{{k}+\mathrm{2}}{{k}!\left({k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{4}\right)} \\ $$$$=\frac{\mathrm{1}}{{k}!\left({k}+\mathrm{2}\right)} \\ $$$$=\frac{{k}+\mathrm{1}}{\left({k}+\mathrm{2}\right)!} \\ $$$$=\frac{{k}+\mathrm{2}−\mathrm{1}}{\left({k}+\mathrm{2}\right)!} \\ $$$$=\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)!}−\frac{\mathrm{1}}{\left({k}+\mathrm{2}\right)!} \\ $$$$ \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}=\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{1}\right)!}−\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)!}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)!} \\ $$$${n}=\mathrm{999} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{999}} {\sum}}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{1001}!}\:\checkmark \\ $$

Commented by Tawa11 last updated on 08/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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