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Question Number 49095 by maxmathsup by imad last updated on 02/Dec/18
$${prove}\:{the}\:{existence}\:{of}\:{n}\:{integrs}\:{naturals}\:{x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,....{x}_{{n}} \:\:\:\:{with}\:{x}_{{i}} \neq{x}_{{j}} {for}\:{i}\neq{j} \\ $$$${and}\:\frac{\mathrm{1}}{{x}_{\mathrm{1}} }\:+\frac{\mathrm{1}}{{x}_{\mathrm{2}} }\:+....+\frac{\mathrm{1}}{{x}_{{n}} }\:=\mathrm{1}\:. \\ $$
Answered by kwonjun1202 last updated on 03/Dec/18
$$\mathrm{1}=\mathrm{0}.\mathrm{5}+\mathrm{0}.\mathrm{25}+\mathrm{0}.\mathrm{125}+\mathrm{0}.\mathrm{0625}+\mathrm{0}.\mathrm{003125}... \\ $$$$=\frac{\mathrm{5}}{\mathrm{10}}\:+\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{10}^{\mathrm{2}} }\:+\frac{\mathrm{5}^{\mathrm{3}} }{\mathrm{10}^{\mathrm{3}} }\:+....+\left(\frac{\mathrm{5}}{\mathrm{10}}\right)^{{n}} \:{reduce}\:{a}\:{fraction}\:=\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$$\therefore\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:=\mathrm{1},\: \\ $$