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}} \\ $$ Answered by john santu last updated on 31/Aug/20…
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Question Number 176365 by Shrinava last updated on 16/Sep/22 $$\mathrm{In}\:\:\bigtriangleup\mathrm{ABC}\:\:\mathrm{the}\:\mathrm{following}\:\mathrm{relationship}\:\mathrm{holds}: \\ $$$$\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} }\:+\:\mathrm{4}\:\underset{\boldsymbol{\mathrm{cyc}}} {\prod}\:\mathrm{cos}\:\mathrm{A}\:\leqslant\:\mathrm{2} \\ $$ Answered by behi834171 last updated on…
Question Number 45283 by peter frank last updated on 11/Oct/18 $$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{x}}=\mathrm{1}+\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\ldots\ldots\ldots \\ $$$$\:\:\:\:\boldsymbol{\mathrm{y}}=\mathrm{1}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\ldots\ldots\ldots \\ $$$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}} \\ $$$$\mathrm{1}+\boldsymbol{\mathrm{ab}}+\boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{3}} \boldsymbol{\mathrm{b}}^{\mathrm{3}} +\ldots\ldots\ldots=\frac{\boldsymbol{\mathrm{xy}}}{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}−\mathrm{1}}…
Question Number 176347 by cortano1 last updated on 16/Sep/22 $$\left(\mathrm{2}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{5}} \left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{5}} \\ $$$$\mathrm{find}\:\mathrm{coefficient}\:\mathrm{x}^{\mathrm{2}} \\ $$ Answered by mr W last updated on 16/Sep/22…
Question Number 110799 by dw last updated on 30/Aug/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{What}\:{is}\:{the}\:{he}\:{minimum}\:{value}\:{of}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\sqrt{\left({y}−{x}\right)^{\mathrm{2}} +\mathrm{25}}+\sqrt{\left({z}−{y}\right)^{\mathrm{2}} +\mathrm{4}}+\sqrt{\left(\mathrm{9}−{z}\right)^{\mathrm{2}} +\mathrm{16}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:\:\left({x},\:{y}\:{and}\:{z}\right)\:\in\:\mathbb{R}\: \\ $$ Terms of Service Privacy Policy…
Question Number 176327 by mathlove last updated on 16/Sep/22 $${p}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} +\sqrt{\mathrm{4}{x}^{\mathrm{2}} }\:\:\: \\ $$$${pq}=\mathrm{4}\:\:\:\:{and}\:\:{p}+{q}=\mathrm{2}\:\:\:{faind} \\ $$$${p}^{{a}_{\mathrm{2}} } +{q}^{{a}_{\mathrm{1}} } =? \\ $$ Terms of Service…