Question Number 88458 by TawaTawa1 last updated on 10/Apr/20 $$\mathrm{Using}\:\mathrm{the}\:\mathrm{principle}\:\mathrm{of}\:\mathrm{mathematical}\:\mathrm{induction}\:\mathrm{to}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\:\:\mathrm{a}_{\mathrm{1}} \:,\:\:\:\mathrm{a}_{\mathrm{2}} \:,\:\:…\:,\:\mathrm{a}_{\mathrm{n}} \:,\:\:\frac{\mathrm{a}_{\mathrm{1}} \:+\:\mathrm{a}_{\mathrm{2}} \:+\:…\:+\:\mathrm{a}_{\mathrm{n}} }{\mathrm{n}}\:\:\:\:\geqslant\:\:\:\sqrt[{\mathrm{n}}]{\mathrm{a}_{\mathrm{1}} \:,\:\:\mathrm{a}_{\mathrm{2}} \:,\:\:…\:,\:\mathrm{a}_{\mathrm{n}} } \\ $$ Terms of…
Question Number 22923 by selestian last updated on 24/Oct/17 Answered by $@ty@m last updated on 24/Oct/17 $${I}_{{n}} =\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{tan}\:^{{n}} {x}\mathrm{sec}\:^{\mathrm{2}} {xdx} \\ $$$${I}_{{n}} =\left[\frac{\mathrm{tan}\:^{{n}+\mathrm{1}}…
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Question Number 88456 by ajfour last updated on 10/Apr/20 Commented by ajfour last updated on 11/Apr/20 $${Find}\:{s}\:{in}\:{terms}\:\:\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}. \\ $$ Commented by Tony Lin last updated…
Question Number 153989 by mathdanisur last updated on 12/Sep/21 $$\mathrm{find}\:\mathrm{all}\:\mathrm{functions}\:\:\mathrm{f}\::\:\mathbb{R}\rightarrow\mathbb{R}\:\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{property}\:\mathrm{that} \\ $$$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\:+\:\mathrm{2y}\:=\:\mathrm{10x}\:+\:\mathrm{f}\left(\mathrm{f}\left(\mathrm{y}\right)-\mathrm{3x}\right)\right. \\ $$$$\mathrm{holds}\:\mathrm{for}\:\mathrm{all}\:\:\mathrm{a};\mathrm{b}\in\mathbb{R} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 153988 by mathdanisur last updated on 12/Sep/21 $$\mathrm{let}\:\:\mathrm{a};\mathrm{b}\:\:\mathrm{be}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers} \\ $$$$\mathrm{such}\:\mathrm{that}\:\:\mathrm{a}+\mathrm{b}=\mathrm{2}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{1}}{\mathrm{a}^{\boldsymbol{\mathrm{n}}} }\:+\:\frac{\mathrm{1}}{\mathrm{b}^{\boldsymbol{\mathrm{n}}} }\:\geqslant\:\mathrm{a}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \:+\:\mathrm{b}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \:\:;\:\:\forall\mathrm{n}\in\mathbb{N}^{\ast} \\ $$ Answered by metamorfose last updated…
Question Number 153977 by mr W last updated on 12/Sep/21 Answered by mr W last updated on 12/Sep/21 $$\frac{{l}}{{a}}=\frac{\mathrm{sin}\:\left(\mathrm{45}+\alpha\right)}{\mathrm{sin}\:\alpha}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}+\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}=\frac{\sqrt{\mathrm{2}}{l}}{{a}}−\mathrm{1} \\ $$$${similarly} \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\beta}=\frac{\sqrt{\mathrm{2}}{l}}{{b}}−\mathrm{1}…
Question Number 88440 by M±th+et£s last updated on 10/Apr/20 $${prove}\:\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{4}^{{k}} −\mathrm{3}^{{k}} }{\mathrm{12}^{{k}} }=\frac{\mathrm{1}}{\mathrm{6}} \\ $$ Commented by Tony Lin last updated on 10/Apr/20…
Question Number 153973 by liberty last updated on 12/Sep/21 Answered by ARUNG_Brandon_MBU last updated on 12/Sep/21 $${A}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}+\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\mathrm{1}}{{n}+\frac{\mathrm{2}}{\mathrm{3}}}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{3}}\left(\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right)=\frac{\pi}{\mathrm{3}}\mathrm{cot}\left(\frac{\pi}{\mathrm{3}}\right)=\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{9}} \\ $$…
Question Number 88438 by M±th+et£s last updated on 10/Apr/20 $$\int\frac{{x}^{\mathrm{5}} +\mathrm{1}}{{x}^{\mathrm{5}} −\mathrm{1}}{dx} \\ $$ Answered by Kunal12588 last updated on 10/Apr/20 $${I}=\int\frac{{x}^{\mathrm{5}} −\mathrm{1}}{{x}^{\mathrm{5}} −\mathrm{1}}{dx}+\mathrm{2}\int\frac{{dx}}{{x}^{\mathrm{5}} −\mathrm{1}}…