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Let-R-n-m-R-n-1-m-R-n-1-m-1-if-R-1-1-6-R-2-1-8-R-3-1-1-and-t-R-4-t-0-a-Find-R-1-21-b-Find-R-1-n-

Question Number 135971 by Raxreedoroid last updated on 17/Mar/21 $$ \\ $$$$\mathrm{Let}\:{R}\left({n},{m}\right)={R}\left({n}−\mathrm{1},{m}\right)+{R}\left({n}−\mathrm{1},{m}−\mathrm{1}\right) \\ $$$$\mathrm{if}\:{R}\left(\mathrm{1},\mathrm{1}\right)=−\mathrm{6},{R}\left(\mathrm{2},\mathrm{1}\right)=\mathrm{8},{R}\left(\mathrm{3},\mathrm{1}\right)=−\mathrm{1} \\ $$$$\mathrm{and}\:\forall{t}\:{R}\left(\mathrm{4},{t}\right)=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\left.{a}\right)\:\mathrm{Find}\:{R}\left(\mathrm{1},\mathrm{21}\right) \\ $$$$ \\…

We-have-released-an-update-with-following-enhancements-A-new-section-quizzes-is-added-You-can-also-request-a-solution-for-quiz-question-and-we-will-provide-a-solution-In-case-you-wan

Question Number 4855 by Tinku Tara last updated on 18/Mar/16 $$\mathrm{We}\:\mathrm{have}\:\mathrm{released}\:\mathrm{an}\:\mathrm{update}\:\mathrm{with}\:\mathrm{following} \\ $$$$\mathrm{enhancements}: \\ $$$$\bullet\:\mathrm{A}\:\mathrm{new}\:\mathrm{section}\:\mathrm{quizzes}\:\mathrm{is}\:\mathrm{added}.\: \\ $$$$\:\:\:\:\mathrm{You}\:\mathrm{can}\:\mathrm{also}\:\mathrm{request}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{for}\:\mathrm{quiz} \\ $$$$\:\:\:\:\mathrm{question}\:\mathrm{and}\:\mathrm{we}\:\mathrm{will}\:\mathrm{provide}\:\mathrm{a}\:\mathrm{solution}. \\ $$$$\:\:\:\:\mathrm{In}\:\mathrm{case}\:\mathrm{you}\:\mathrm{want}\:\mathrm{to}\:\mathrm{start}\:\mathrm{a}\:\mathrm{discussion}\:\mathrm{you} \\ $$$$\:\:\:\:\mathrm{can}\:\mathrm{post}\:\mathrm{the}\:\mathrm{same}\:\mathrm{question}\:\mathrm{to}\:\mathrm{forum}\:\mathrm{directly} \\ $$$$\:\:\:\:\mathrm{from}\:\mathrm{quiz}.…

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Question Number 70360 by Hassen_Timol last updated on 03/Oct/19 $$\underset{{n}\rightarrow+\infty} {\mathrm{l}im}\:\:\:\:\frac{\sqrt{{n}\:+\:\mathrm{1}\:}−\:{n}}{\:\sqrt{{n}\:+\:\mathrm{1}}\:+\:{n}}\:\:=\:\:? \\ $$ Answered by mind is power last updated on 03/Oct/19 $$\frac{\sqrt{{n}+\mathrm{1}}−{n}}{\:\sqrt{{n}+\mathrm{1}}+{n}}=\frac{{n}\left(\sqrt{\frac{{n}+\mathrm{1}}{{n}^{\mathrm{2}} }}−\mathrm{1}\right)}{{n}\left(\sqrt{\frac{{n}+\mathrm{1}}{{n}^{\mathrm{2}} }}+\mathrm{1}\right)}=\frac{\sqrt{\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}^{\mathrm{2}}…