Question Number 13102 by tawa tawa last updated on 14/May/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{nth}\:\mathrm{term}\:\::\:\:\mathrm{1}^{\mathrm{6}} \:+\:\mathrm{2}^{\mathrm{6}} \:+\:\mathrm{3}^{\mathrm{6}} \:+\:\mathrm{4}^{\mathrm{6}} \:+\:…\:+\:\mathrm{n}^{\mathrm{6}} \\ $$ Commented by tawa tawa last updated on 14/May/17…
Question Number 78567 by TawaTawa last updated on 18/Jan/20 Commented by mr W last updated on 18/Jan/20 Commented by mr W last updated on 18/Jan/20…
Question Number 425 by 123456 last updated on 25/Jan/15 $${x}^{\mathrm{2}} +{y}^{\mathrm{3}} ={z}^{\mathrm{4}} \:\mathrm{have}\:\mathrm{integer}\:\boldsymbol{\mathrm{solutions}}? \\ $$ Answered by prakash jain last updated on 02/Jan/15 $$\mathrm{Assuming}\:\mathrm{we}\:\mathrm{are}\:\mathrm{looking}\:\mathrm{for}\:+\mathrm{ve} \\…
Question Number 131484 by bemath last updated on 05/Feb/21 $$\mathrm{Express}\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{49}} {\sum}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}+\sqrt{\mathrm{n}^{\mathrm{2}} −\mathrm{1}}}}\:\mathrm{as}\:{a}+{b}\sqrt{\mathrm{2}} \\ $$$$\mathrm{for}\:\mathrm{some}\:\mathrm{integers}\:{a}\:\mathrm{and}\:{b} \\ $$ Answered by benjo_mathlover last updated on 05/Feb/21 Answered…
Question Number 403 by 123456 last updated on 25/Jan/15 $$\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2005}} }\:\mathrm{have}\:\mathrm{how}\:\mathrm{many}\:\mathrm{digits}\:\mathrm{in}\:\mathrm{periodic}\:\mathrm{part}? \\ $$ Answered by prakash jain last updated on 30/Dec/14 $$\mathrm{3}^{\mathrm{2003}} \:\mathrm{digits}\:\mathrm{will}\:\mathrm{be}\:\mathrm{present}\:\mathrm{in}\:\mathrm{periodic}\:\mathrm{part}. \\ $$$$\mathrm{10}^{{n}}…
Question Number 400 by 123456 last updated on 25/Jan/15 $$\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{20}} }\:\mathrm{have}\:\mathrm{how}\:\mathrm{many}\:\mathrm{digits}\:\mathrm{in}\:\mathrm{periodic}\:\mathrm{part}? \\ $$ Answered by prakash jain last updated on 29/Dec/14 $$\mathrm{3}^{\mathrm{20}} \:\mathrm{is}\:\mathrm{not}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{2}\:\mathrm{or}\:\mathrm{5}.\:\mathrm{So} \\ $$$$\mathrm{periodicity}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\mathrm{equation}…
Question Number 347 by 123456 last updated on 23/Dec/14 $$\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}−\frac{\mathrm{4}}{\mathrm{5}−\ddots}}\overset{?} {=}\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 348 by 123456 last updated on 23/Dec/14 $${a}_{{n}} =\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}−\frac{\mathrm{4}}{\mathrm{5}−{n}}},{n}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$${a}_{{n}} =\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}−\frac{\mathrm{4}}{\mathrm{5}−\frac{\mathrm{6}}{{n}}}},{n}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} \overset{?} {=}\mathrm{1} \\ $$ Commented by 123456 last…
Question Number 215 by 123456 last updated on 25/Jan/15 $$\mathrm{evaluate} \\ $$$$\mathrm{1}+\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{4}^{\mathrm{5}} } \left(\mathrm{mod}\:\mathrm{10}\right) \\ $$ Answered by prakash jain last updated on 16/Dec/14…
Question Number 187 by 123456 last updated on 25/Jan/15 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{integer}\:{x},{y} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{9}{y}^{\mathrm{2}} =\mathrm{6} \\ $$ Answered by ghosea last updated on 16/Dec/14 $$\left(\mathrm{2}{x}−\mathrm{3}{y}\right)\left(\mathrm{2}{x}+\mathrm{3}{y}\right)=\mathrm{2}\centerdot\mathrm{3}\:{or}\:\mathrm{1}.\mathrm{6}\:{or}\:−\mathrm{1},−\mathrm{6}\:{or}−\mathrm{2},−\mathrm{3} \\…