Question Number 176423 by Shrinava last updated on 18/Sep/22 Answered by a.lgnaoui last updated on 21/Sep/22 $$\:\frac{\mathrm{3}\pi}{{x}}=\left(\:\frac{\mathrm{4}\pi}{{x}}−\frac{\pi}{{x}}\right);\:\:\:\frac{\mathrm{5}\pi}{{x}}=\left(\frac{\mathrm{4}\pi}{{x}}+\frac{\pi}{{x}}\right);\:\:\frac{\mathrm{4}\pi}{{x}}=\mathrm{2}\left(\frac{\mathrm{2}\pi}{{x}}\right)\:\:\: \\ $$$$\mathrm{cos}\:\frac{\mathrm{3}\pi}{{x}}+\mathrm{cos}\:\frac{\mathrm{5}\pi}{{x}}=\mathrm{cos}\left(\:\frac{\mathrm{4}\pi}{{x}}−\frac{\pi}{{x}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{{x}}+\frac{\pi}{{x}}\right)=\mathrm{2cos}\:\frac{\pi}{{x}}\mathrm{cos}\:\frac{\mathrm{4}\pi}{{x}} \\ $$$$ \\ $$$$\mathrm{cos}\:\frac{\pi}{{x}}+\mathrm{2cos}\:\frac{\pi}{{x}}\mathrm{cos}\:\frac{\mathrm{4}\pi}{{x}}=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\frac{\mathrm{2}\pi}{{x}}+\mathrm{cos}\:\frac{\mathrm{4}\pi}{{x}} \\ $$$$\mathrm{cos}\:\frac{\pi}{{x}}\left(\mathrm{1}+\mathrm{2cos}\:\frac{\mathrm{4}\pi}{{x}}\right)=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\frac{\mathrm{4}\pi}{{x}}+\mathrm{cos}\:\frac{\mathrm{2}\pi}{{x}}…
Question Number 110879 by Khanacademy last updated on 31/Aug/20 Commented by Khanacademy last updated on 31/Aug/20 $$\boldsymbol{{PLEASE}} \\ $$$$ \\ $$ Terms of Service Privacy…
Question Number 45327 by Necxx last updated on 12/Oct/18 Answered by rahul 19 last updated on 12/Oct/18 $${x}=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}+\frac{\mathrm{1}}{\mathrm{1}+{x}}} \\ $$$$\Rightarrow{x}=\mathrm{1}+\frac{\mathrm{1}+{x}}{\mathrm{4}+\mathrm{3}{x}} \\ $$$$\Rightarrow\mathrm{4}{x}+\mathrm{3}{x}^{\mathrm{2}} =\mathrm{5}+\mathrm{4}{x} \\ $$$$\Rightarrow{x}=\:\underset{−}…
Question Number 176393 by BaliramKumar last updated on 18/Sep/22 $${show}\:{that} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…………….\:\infty\:=\:\frac{−\mathrm{1}}{\mathrm{8}} \\ $$ Answered by BaliramKumar last updated on 19/Sep/22 $$\mathrm{S}\:=\:\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+\mathrm{8}+\mathrm{9}+\mathrm{10}+……..\infty \\ $$$$\mathrm{S}\:=\:\mathrm{1}+\underset{\mathrm{9}} {\underbrace{\mathrm{2}+\mathrm{3}+\mathrm{4}}}+\underset{\mathrm{18}}…
Question Number 176391 by cortano1 last updated on 18/Sep/22 Commented by mr W last updated on 18/Sep/22 $${f}\left({x}\right)={x}^{\mathrm{3}} \\ $$ Answered by TheHoneyCat last updated…
Question Number 176384 by Ari last updated on 17/Sep/22 $${Find}\:{the}\:{constant}\:{term}\:{in}\:{the} \\ $$$${expansion}\:{of}\:{the}\:{expression} \\ $$$$\left(\mathrm{2}+\mathrm{3}{x}\right)^{\mathrm{3}} \left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right)^{\mathrm{4}} \\ $$ Answered by mr W last updated on 17/Sep/22…
Question Number 45316 by Tawa1 last updated on 11/Oct/18 $$\mathrm{I}\:\mathrm{heard}\:\mathrm{this}\:\mathrm{sum}\:\mathrm{can}\:\mathrm{be}\:\mathrm{close}\:\mathrm{using}\:\mathrm{the}\:\mathrm{Digamma}\:\mathrm{function}. \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{use}\:\mathrm{it}.\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{it}.\:\:\: \\ $$$$\:\:\:\mathrm{sum}\:\mathrm{of}\:\mathrm{nth}\:\mathrm{term}:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}\:+\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{8}.\mathrm{9}}\:+\:…\: \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18 $${T}_{{n}} =\frac{\mathrm{1}}{\left\{\mathrm{1}+\left({n}−\mathrm{1}\right)\mathrm{3}\right\}\left\{\mathrm{2}+\left({n}−\mathrm{1}\right)\mathrm{3}\right\}\left\{\mathrm{3}+\left({n}−\mathrm{1}\right)\mathrm{3}\right\}}…
Question Number 176387 by cortano1 last updated on 19/Sep/22 $$\:\:\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\:=\:\mathrm{1}\:\Rightarrow\frac{\left[\mathrm{x}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }\right]^{\mathrm{3}} −\mathrm{1}}{\mathrm{x}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }}\:=? \\ $$ Commented by mr W last updated…
Question Number 110843 by dw last updated on 31/Aug/20 $$\left({x}+\mathrm{1}\right)^{\left({x}+\mathrm{1}\right)} =\sqrt{\mathrm{2}}\:\:\:\:\:\:{find}\:{all}\:{values}\:{of}\:{x} \\ $$$$\left({Please}\:{step}\:{by}\:{step}\right) \\ $$ Answered by john santu last updated on 31/Aug/20 $$\rightarrow\mathrm{ln}\:\left({x}+\mathrm{1}\right)^{\left({x}+\mathrm{1}\right)} =\mathrm{ln}\:\left(\sqrt{\mathrm{2}}\right)…
Question Number 110845 by bemath last updated on 31/Aug/20 $$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{10}\right)^{\mathrm{x}^{\mathrm{3}} −\mathrm{9x}} \:=\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{10}\right)^{\mathrm{3x}^{\mathrm{2}} −\mathrm{8x}} \\ $$ Answered by john santu last updated on 31/Aug/20…