Question Number 28743 by Cheyboy last updated on 29/Jan/18 $${find}\:{the}\:{next}\:\mathrm{4}\:{term}\:{and}\:{the} \\ $$$${n}^{{th}} \:{term} \\ $$$$\mathrm{1},\mathrm{2},\mathrm{5},\mathrm{26}……. \\ $$ Commented by Tinkutara last updated on 29/Jan/18 $${a}_{{n}}…
Question Number 159750 by greg_ed last updated on 20/Nov/21 $$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{that}}\: \\ $$$$\:\sqrt{\mathrm{689}×\mathrm{690}×\mathrm{691}+\mathrm{1}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{naturel}}\:\boldsymbol{\mathrm{number}}. \\ $$ Commented by Rasheed.Sindhi last updated on 20/Nov/21 $${Pl}\:{check}\:{the}\:{question},\:\sqrt{\mathrm{689}×\mathrm{690}×\mathrm{691}+\mathrm{1}}\: \\ $$$${is}\:{not}\:{a}\:{natural}\:{number}. \\…
Question Number 28607 by abdo imad last updated on 27/Jan/18 $${simplify}\:\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\mathrm{3}^{{k}} \:{sin}^{\mathrm{3}} \left(\frac{\alpha}{\mathrm{3}^{{k}+\mathrm{1}} }\right)\:. \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 94097 by Rio Michael last updated on 16/May/20 $$\mathrm{find}\:\mathrm{all}\:\mathrm{integers}\:{n}\:\:\mathrm{for}\:\mathrm{which}\:\:\mathrm{13}\:\mid\mathrm{4}\left({n}^{\mathrm{2}} +\mathrm{1}\right). \\ $$ Commented by Rasheed.Sindhi last updated on 17/May/20 $${This}\:{s}\:{equivalent}\:{to}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{13}\:\mid\left({n}^{\mathrm{2}} +\mathrm{1}\right)…
Question Number 159556 by mnjuly1970 last updated on 18/Nov/21 $$ \\ $$$$\:\:{prove}\:{that}: \\ $$$$ \\ $$$$\:\:\:\:\mathrm{2}\nmid\:{a}\:\Rightarrow\:\mathrm{240}\mid\:{a}^{\:\mathrm{5}} \:−\:{a}\:\:\:\:\: \\ $$$$ \\ $$ Answered by floor(10²Eta[1]) last…
Question Number 28449 by A1B1C1D1 last updated on 25/Jan/18 Answered by $@ty@m last updated on 26/Jan/18 $${there}\:{are}\:{infinitely}\:{many}\:{solutions}: \\ $$$$\left(\mathrm{0},\pm\mathrm{1}\right) \\ $$$$\left(\pm\mathrm{1},\mathrm{0}\right) \\ $$$$\left(\pm\frac{\mathrm{3}}{\mathrm{5}},\pm\frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$$$\left(\pm\frac{\mathrm{5}}{\mathrm{13}},\pm\frac{\mathrm{12}}{\mathrm{13}}\right)…
Question Number 93859 by M±th+et+s last updated on 15/May/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)} \\ $$ Answered by Ar Brandon last updated on 15/May/20 $$\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\Rightarrow\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}=\underset{\mathrm{n}=\mathrm{1}}…
Question Number 28289 by neel1974 last updated on 23/Jan/18 $${find}\:{the}\:{loss}\:{amt}\:{when}\:{cost}\:{price} \\ $$$${and}\:\%\:{of}\:{loss}\:{given} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 93791 by peter frank last updated on 15/May/20 Commented by peter frank last updated on 15/May/20 $${iii} \\ $$ Terms of Service Privacy…
Question Number 93782 by M±th+et+s last updated on 14/May/20 $${find}\:\frac{{a}}{{b}} \\ $$$${a}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{k}}} \\ $$$${b}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{k}+\mathrm{1}}} \\ $$$$ \\ $$ Terms of Service…