Question Number 6289 by sanusihammed last updated on 22/Jun/16 Commented by FilupSmith last updated on 22/Jun/16 $${a}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{6}}+…+\frac{\mathrm{2015}}{\mathrm{2016}} \\ $$$${b}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{4}}{\mathrm{6}}×…×\frac{\mathrm{2015}}{\mathrm{2016}} \\ $$$$\frac{{a}}{{b}}+\frac{{b}}{{a}}=?? \\ $$$$ \\ $$$${a}=\underset{{n}=\mathrm{1}}…
Question Number 71790 by jatin123 last updated on 20/Oct/19 Answered by $@ty@m123 last updated on 20/Oct/19 $$=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{4}}×…..×\frac{\mathrm{98}}{\mathrm{99}}×\frac{\mathrm{99}}{\mathrm{100}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{100}}=\frac{\mathrm{1}}{\mathrm{50}} \\ $$ Commented by jatin123 last…
Question Number 137324 by greg_ed last updated on 01/Apr/21 $$\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{evaluate}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{one}}\:: \\ $$$$\mathrm{P}\:=\:\left(\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{1958}}\right)\left(\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{1959}}\right)\left(\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{1960}}\right)…\left(\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{2017}}\right)\left(\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{2018}}\right)\left(\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{2019}}\right) \\ $$$$\boldsymbol{\mathrm{P}}\:=\:?\: \\ $$ Answered by som(math1967) last updated on 01/Apr/21 $${P}=\left(\frac{\mathrm{1959}}{\mathrm{1958}}\right)\left(\frac{\mathrm{1960}}{\mathrm{1959}}\right)\left(\frac{\mathrm{1961}}{\mathrm{1960}}\right)..\left(\frac{\mathrm{2019}}{\mathrm{2018}}\right)\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right) \\…
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Question Number 71717 by peter frank last updated on 19/Oct/19 Commented by mathmax by abdo last updated on 19/Oct/19 $${let}\:{I}=\int\:\frac{{dx}}{{cos}\left({x}−{a}\right){cos}\left({x}−{b}\right)}\:\Rightarrow{I}=\int\:\:\frac{\mathrm{2}{dx}}{{cos}\left(\mathrm{2}{x}−{a}−{b}\right)+{cos}\left({a}−{b}\right)} \\ $$$$=_{\mathrm{2}{x}−\left({a}+{b}\right)={t}} \:\:\:\:\int\:\:\frac{{dt}}{{cost}\:+\lambda}\:\:{with}\:\lambda={cos}\left({a}−{b}\right)\:{changement} \\ $$$${tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}\:{give}\:\int\:\:\:\frac{{dt}}{{cost}\:+\lambda}\:=\int\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{u}^{\mathrm{2}}…
Question Number 6166 by sanusihammed last updated on 16/Jun/16 $${If}\:{the}\:{sum}\:{of}\:{the}\:{first}\:{nth}\:{terms}\:{of}\:{a}\:{sequence}\:{is}\:{given}\:{by}\: \\ $$$${S}_{{n}\:} \:=\:\:\mathrm{9}\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{3}^{{n}} \:}\right) \\ $$$$\left({a}\right)\:{find}\:{the}\:{first}\:{and}\:{the}\:{second}\:{term}\:{of}\:{the}\:{sequence} \\ $$$$\left({b}\right)\:{find}\:{the}\:{nth}\:{term}\:{of}\:{the}\:{sequence} \\ $$$$\left({c}\right)\:{show}\:{that}\:{the}\:{sequence}\:{is}\:{a}\:{GP}\:{and}\:{find}\:{it}\:{common}\:{ratio} \\ $$$$ \\ $$$${please}\:{help}. \\…
Question Number 6158 by enigmeyou last updated on 16/Jun/16 $$\frac{−\mathrm{1}}{\mathrm{1048576}}=\frac{\mathrm{1}}{\mathrm{8}}×\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$${find}\:{n}\:? \\ $$ Answered by Rasheed Soomro last updated on 16/Jun/16 $$\frac{−\mathrm{1}}{\mathrm{1048576}}=\frac{\mathrm{1}}{\mathrm{8}}×\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$$\frac{−\mathrm{8}}{\mathrm{1048576}}=\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)^{{n}}…
Question Number 71695 by peter frank last updated on 18/Oct/19 Answered by MJS last updated on 19/Oct/19 $$\int\frac{\sqrt{\mathrm{tan}\:{x}}}{\mathrm{1}+\sqrt{\mathrm{tan}\:{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{tan}\:{x}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{\mathrm{tan}\:{x}}\mathrm{cos}^{\mathrm{2}} \:{x}\:{dt}=\frac{\mathrm{2}{t}}{{t}^{\mathrm{4}} +\mathrm{1}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{4}}…
Question Number 137206 by JulioCesar last updated on 31/Mar/21 Answered by bemath last updated on 31/Mar/21 $$\mathrm{by}\:\mathrm{parts}\:\begin{cases}{\mathrm{u}=\mathrm{ln}\:\left(\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}\right) \:\mathrm{du}=\frac{\mathrm{2}}{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)}\mathrm{dx}}\\{\mathrm{v}\:=\:\mathrm{x}}\end{cases} \\ $$$$\mathrm{I}\:=\:\mathrm{x}\:\mathrm{ln}\:\left(\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}\right)−\int\:\frac{\mathrm{2x}}{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)}\mathrm{dx} \\ $$$$\mathrm{I}=\mathrm{x}\:\mathrm{ln}\:\left(\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}\right)−\left[\int\:\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}\mathrm{dx}+\int\:\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}\mathrm{dx}\:\right] \\ $$$$\mathrm{I}=\:\mathrm{x}\:\mathrm{ln}\:\left(\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}\right)−\mathrm{ln}\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\:+\:\mathrm{C}…
Question Number 6087 by sanusihammed last updated on 12/Jun/16 $${Given}\:{the}\:{terms}\:\:\:\:{a}_{{k}} \:\:\:{and}\:\:{a}_{{k}\:+\:{r}\:} \:\:{of}\:{an}\:{A}.{P} \\ $$$${find}\:\:{a}_{\mathrm{1}} \:\:{and}\:\:{a}_{{k}\:+\:{r}} \:\:{in}\:{terms}\:{of}\:\:{a}_{{k}\:} \:{and}\:\:{a}_{{k}\:+\:{r}} \\ $$ Commented by Rasheed Soomro last updated…