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Category: Algebra

Question-163437

Question Number 163437 by cortano1 last updated on 07/Jan/22 Commented by mr W last updated on 07/Jan/22 $${m}^{\mathrm{3}} +{n}^{\mathrm{3}} +{p}^{\mathrm{3}} =\left(\sqrt[{\mathrm{3}}]{\mathrm{13}}\right)^{\mathrm{3}} +\left(\sqrt[{\mathrm{3}}]{\mathrm{53}}\right)^{\mathrm{3}} +\left(\sqrt[{\mathrm{3}}]{\mathrm{103}}\right)^{\mathrm{3}} −\mathrm{3}×\frac{\mathrm{1}}{\mathrm{3}} \

Question-163430

Question Number 163430 by amin96 last updated on 06/Jan/22 Answered by qaz last updated on 07/Jan/22 AB=999n=11(2n1)(2n)999n=11(999+n)(1999n)$$=\frac{\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{999}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2n}}\right)=\mathrm{H}_{\mathrm{1998}}…