Question Number 189254 by yaslm last updated on 13/Mar/23 Answered by manxsol last updated on 14/Mar/23 $$\begin{vmatrix}{{a}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{{a}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{}&{}&{{a}}&{}&{}&{}&{\mathrm{1}}\\{}&{}&{}&{.}&{}&{}&{}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{{a}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{{a}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{{a}}\end{vmatrix} \\ $$$$−{fn}+{f}_{{n}−\mathrm{1}} \\ $$$$−{fn}+{f}_{{n}−\mathrm{2}} \\ $$$$. \\ $$$$−{f}_{{n}}…
Question Number 189248 by 073 last updated on 13/Mar/23 Answered by mr W last updated on 13/Mar/23 $$\underset{{n}=\mathrm{1}} {\overset{\mathrm{1012}} {\sum}}\left[\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} −\left(\mathrm{2}{n}\right)^{\mathrm{3}} \right] \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{1012}}…
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Question Number 189223 by sciencestudentW last updated on 13/Mar/23 $${who}\:{did}\:{discoer}\:{the}\:{light}'{s}\:{speed}\:{and} \\ $$$${by}\:{which}\:{method}? \\ $$ Commented by mr W last updated on 13/Mar/23 $${are}\:{google}\:{and}\:{wikipedia}\:{etc}.\:{forbidden} \\ $$$${in}\:{your}\:{country}?…
Question Number 123681 by ZiYangLee last updated on 27/Nov/20 $$\mathrm{How}\:\mathrm{many}\:\mathrm{3}-\mathrm{digits}\:\mathrm{positive}\:\mathrm{integers} \\ $$$$\mathrm{are}\:\mathrm{there}\:\mathrm{such}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\: \\ $$$$\mathrm{is}\:\mathrm{10}? \\ $$ Answered by mr W last updated on 27/Nov/20 $${abc}\:{with}…
Question Number 123661 by weltr last updated on 27/Nov/20 $${find}\:{sum}\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\sqrt{\mathrm{2}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:+\:\sqrt{\mathrm{3}}}\:+\:…\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2019}}\:+\:\sqrt{\mathrm{2020}}} \\ $$ Answered by Dwaipayan Shikari last updated on 27/Nov/20 $$\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}}=\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}} \\ $$$${sum}\:\:=\:\sqrt{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{4}}−\sqrt{\mathrm{3}}+…+\sqrt{\mathrm{2020}}−\sqrt{\mathrm{2019}}…
Question Number 189183 by pascal889 last updated on 12/Mar/23 Answered by Rasheed.Sindhi last updated on 13/Mar/23 $${a}:\:{first}\:{term}\:{of}\:{AP}, \\ $$$${g}:\:{first}\:{term}\:{of}\:{GP} \\ $$$${a}_{\mathrm{1}} +{g}_{\mathrm{1}} ={a}+{g}=\mathrm{8}\Rightarrow{a}=\mathrm{8}−{g} \\ $$$${a}_{\mathrm{3}}…
Question Number 58113 by MJS last updated on 17/Apr/19 $$\mathrm{once}\:\mathrm{sgain}:\:\mathrm{it}'\mathrm{s}\:\mathrm{boring}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{questions}\:\mathrm{of} \\ $$$$\mathrm{minor}\:\mathrm{complexity}.\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{to},\:\mathrm{we}\:\mathrm{do} \\ $$$$\mathrm{it}\:\mathrm{to}\:\mathrm{help}\:\mathrm{unexperienced}\:\mathrm{people}\:\mathrm{to}\:\mathrm{grow}. \\ $$$$\mathrm{you}\:\mathrm{could}\:\mathrm{at}\:\mathrm{least}\:\mathrm{type}\:“\mathrm{thanks}''.\:\mathrm{otherwise} \\ $$$$\mathrm{you}\:\mathrm{might}\:\mathrm{be}\:\mathrm{ignored}\:\mathrm{after}\:\mathrm{a}\:\mathrm{while}… \\ $$ Commented by mr W last…
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Question Number 123643 by aurpeyz last updated on 27/Nov/20 $${find}\:{the}\:{area}\:{bounded}\:{by}\:{the}\:{curve} \\ $$$${x}^{\mathrm{5}} =\mathrm{5}{y}\:{and}\:{straight}\:{line}\:{x}=\mathrm{3}{y}−\mathrm{4} \\ $$ Commented by benjo_mathlover last updated on 27/Nov/20 $${area}\:=\:\mathrm{0} \\ $$…