Question Number 147877 by Tawa11 last updated on 24/Jul/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{close}\:\mathrm{form}. \\ $$$$\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}\:\:\:+\:\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{8}.\mathrm{9}}\:\:\:+\:\:\:…\:\:\: \\ $$ Commented by Tinku Tara last updated on 24/Jul/21 $$\mathrm{Welcome}\:\mathrm{back}.\:\mathrm{Tawa}. \\ $$🙂🙂…
Question Number 147882 by mathdanisur last updated on 24/Jul/21 $${if}\:\:\:\sqrt[{\mathrm{3}}]{\mathrm{6}\sqrt{\mathrm{3}}\:+\:\mathrm{10}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{6}\sqrt{\mathrm{3}}\:−\:\mathrm{10}}\:=\:\boldsymbol{{x}} \\ $$$${find}\:\:\:\frac{{x}^{\mathrm{3}} }{\mathrm{10}−\mathrm{3}{x}}\:=\:? \\ $$ Answered by mr W last updated on 24/Jul/21 $${a}=\sqrt[{\mathrm{3}}]{\mathrm{6}\sqrt{\mathrm{3}}\:+\:\mathrm{10}}\:,{b}=\:\sqrt[{\mathrm{3}}]{\mathrm{6}\sqrt{\mathrm{3}}\:−\:\mathrm{10}} \\…
Question Number 147878 by Tawa11 last updated on 24/Jul/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\:\mathrm{n}\:\:\mathrm{term}:\:\:\:\:\:\mathrm{1}.\mathrm{2}.\mathrm{3}\:\:+\:\:\mathrm{4}.\mathrm{5}.\mathrm{6}\:\:+\:\:\mathrm{7}.\mathrm{8}.\mathrm{9}\:\:+\:\:… \\ $$ Answered by Olaf_Thorendsen last updated on 24/Jul/21 $$\mathrm{S}\:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{3}{k}−\mathrm{2}\right)\left(\mathrm{3}{k}−\mathrm{1}\right)\left(\mathrm{3}{k}\right) \\ $$$$\mathrm{S}\:=\:\mathrm{3}\underset{{k}=\mathrm{1}} {\overset{{n}}…
Question Number 16794 by tawa tawa last updated on 26/Jun/17 Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/Jun/17 $${c}−{x}={t}^{\mathrm{3}} ,{x}−{a}={s}^{\mathrm{3}} \Rightarrow{x}=\frac{{s}^{\mathrm{3}} −{t}^{\mathrm{3}} }{\mathrm{2}}+\frac{{a}+{c}}{\mathrm{2}} \\ $$$${a}+{c}={c}−\mathrm{2}+{c}=\mathrm{2}\left({c}−\mathrm{1}\right)=\mathrm{2}{b},{c}−{a}=\mathrm{2} \\…
Question Number 82307 by Power last updated on 20/Feb/20 Commented by Tony Lin last updated on 20/Feb/20 $$\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)} \\ $$$$=\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\…
Question Number 147842 by mathdanisur last updated on 23/Jul/21 $${x}\:;\:{y}\:;\:{z}\:>\:\mathrm{0} \\ $$$$\begin{cases}{{x}+{y}^{\mathrm{2}} +{z}^{\mathrm{3}} \:=\:\mathrm{3}}\\{{y}+{z}^{\mathrm{2}} +{x}^{\mathrm{3}} \:=\:\mathrm{3}}\\{{z}+{x}^{\mathrm{2}} +{y}^{\mathrm{3}} \:=\:\mathrm{3}}\end{cases}\:\:\Rightarrow\:\:{x}\:;\:{y}\:;\:{z}\:=\:? \\ $$ Commented by mr W last…
Question Number 147835 by peter frank last updated on 23/Jul/21 $${prove}\:{that}\: \\ $$$$\:\:\int\mathrm{cos}\:\mathrm{2}\theta{log}\left(\frac{\mathrm{cos}\:\theta+\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta−\mathrm{sin}\:\theta}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}\theta{log}\left[\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\theta\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\left(\mathrm{cos}\:\mathrm{2}\theta\right)\right. \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 147809 by mathdanisur last updated on 23/Jul/21 $$\frac{\boldsymbol{{sin}}^{\mathrm{6}} \left(\boldsymbol{\alpha}\right)\:+\:\boldsymbol{{cos}}^{\mathrm{6}} \left(\boldsymbol{\alpha}\right)\:-\:\mathrm{1}}{\boldsymbol{{sin}}^{\mathrm{4}} \left(\boldsymbol{\alpha}\right)\:-\:\boldsymbol{{sin}}^{\mathrm{2}} \left(\boldsymbol{\alpha}\right)}\:=\:? \\ $$ Answered by gsk2684 last updated on 23/Jul/21 $$\left(\mathrm{sin}\:^{\mathrm{2}} \alpha\right)^{\mathrm{3}}…
Question Number 147810 by mathdanisur last updated on 23/Jul/21 $$\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)\:\centerdot\:\boldsymbol{{cos}}\left(\mathrm{3}\boldsymbol{{x}}\right)=\boldsymbol{{cos}}\left(\mathrm{5}\boldsymbol{{x}}\right)\:\centerdot\:\boldsymbol{{cos}}\left(\mathrm{7}\boldsymbol{{x}}\right) \\ $$$$\boldsymbol{{x}}\:=\:? \\ $$ Answered by gsk2684 last updated on 23/Jul/21 $$\mathrm{2}\:\mathrm{cos}\:{x}\:\mathrm{cos}\:\mathrm{3}{x}\:=\:\mathrm{2}\:\mathrm{cos}\:\mathrm{5}{x}\:\mathrm{cos}\:\mathrm{7}{x} \\ $$$$\mathrm{cos}\:\left(\mathrm{3}{x}+{x}\right)+\mathrm{cos}\:\left(\mathrm{3}{x}−{x}\right)=\mathrm{cos}\:\left(\mathrm{7}{x}+\mathrm{5}{x}\right)+\mathrm{cos}\:\left(\mathrm{7}{x}−\mathrm{5}{x}\right) \\…