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Question Number 162265 by HongKing last updated on 28/Dec/21 $$\mathrm{Calculate}:\:\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{H}_{\mathrm{2}\boldsymbol{\mathrm{k}}} \:\left(-\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}-\mathrm{1}} }{\mathrm{2k}\:+\:\mathrm{1}} \\ $$$$\mathrm{where},\:\mathrm{H}_{\boldsymbol{\mathrm{n}}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{n}-\mathrm{th}\:\mathrm{harmonic}\:\mathrm{number} \\ $$ Terms of Service Privacy Policy Contact:…
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Question Number 31194 by U Htay KyawMyint last updated on 03/Mar/18 $${Find}\:{the}\:{remainder}\:{when}\:{x}^{\mathrm{203}} −\mathrm{1} \\ $$$${is}\:{divided}\:{by}\:{x}^{\mathrm{4}} −\mathrm{1}. \\ $$ Commented by 6123 last updated on 04/Mar/18…
Question Number 162264 by HongKing last updated on 28/Dec/21 $$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\:\mathrm{dxdydz}\:=\:? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 162261 by mr W last updated on 28/Dec/21 $${find}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{5}^{{n}} −\mathrm{1}}=? \\ $$$${or}\:{generally} \\ $$$$\Phi\left({k}\right)=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{k}^{{n}} −\mathrm{1}}=?\:{with}\:{k}\in{N},\:{k}\geqslant\mathrm{2} \\ $$ Commented by…
Question Number 96712 by john santu last updated on 04/Jun/20 Answered by bobhans last updated on 04/Jun/20 $$\mathrm{since}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{on}\:\left(\mathrm{0},+\infty\right),\:\mathrm{from} \\ $$$$\begin{cases}{\mathrm{2}{a}^{\mathrm{2}} +{a}+\mathrm{1}=\mathrm{2}\left({a}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{8}}>\mathrm{0}}\\{\mathrm{3}{a}^{\mathrm{2}} −\mathrm{4}{a}+\mathrm{1}=\left(\mathrm{3}{a}−\mathrm{1}\right)\left({a}−\mathrm{1}\right)>\mathrm{0}}\end{cases} \\ $$$$\mathrm{we}\:\mathrm{get}\:{a}<\frac{\mathrm{1}}{\mathrm{3}}\:\cup\:{a}>\mathrm{1}…\left({i}\right)…
Question Number 96715 by bobhans last updated on 04/Jun/20 $$\mathrm{find}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{equation} \\ $$$${x}^{\mathrm{5}} +{x}^{\mathrm{4}} +\mathrm{1}\:=\:\mathrm{0} \\ $$ Answered by bemath last updated on 04/Jun/20 $$\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{3}}…
Question Number 162187 by mathlove last updated on 27/Dec/21 $${x}^{{y}} =\mathrm{9} \\ $$$$\sqrt[{{y}}]{\mathrm{32}{y}}=\mathrm{2}{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\left({x},{y}\right)=? \\ $$$${sulotion}\:=? \\ $$$$ \\ $$ Commented by mr W last…
Question Number 96650 by bobhans last updated on 03/Jun/20 $$\mathrm{solve}\:\mathrm{2}\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{2y}−\mathrm{1}}\:=\:\mathrm{y}^{\mathrm{3}} +\mathrm{1} \\ $$ Answered by john santu last updated on 03/Jun/20 $$\sqrt[{\mathrm{3}\:\:}]{\mathrm{2y}−\mathrm{1}}\:=\:\frac{\mathrm{y}^{\mathrm{3}} +\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{f}^{−\mathrm{1}}…