Question Number 103673 by bobhans last updated on 16/Jul/20 $$\underset{{k}=\mathrm{1}} {\overset{\mathrm{4095}} {\sum}}\frac{\mathrm{1}}{\left(\sqrt{{k}}+\sqrt{{k}+\mathrm{1}}\right)\left(\sqrt[{\mathrm{4}}]{{k}}+\sqrt[{\mathrm{4}}]{{k}+\mathrm{1}}\right)}\:? \\ $$ Answered by bramlex last updated on 16/Jul/20 $$\frac{\mathrm{1}}{\left(\sqrt{\mathrm{k}}+\sqrt{\mathrm{k}+\mathrm{1}}\right)\left(\sqrt[{\mathrm{4}}]{\mathrm{k}}+\sqrt[{\mathrm{4}}]{\mathrm{k}+\mathrm{1}}\right)}\:×\frac{\sqrt[{\mathrm{4}}]{\mathrm{k}+\mathrm{1}}−\sqrt[{\mathrm{4}}]{\mathrm{k}}}{\:\sqrt[{\mathrm{4}}]{\mathrm{k}+\mathrm{1}}−\sqrt[{\mathrm{4}}]{\mathrm{k}}}= \\ $$$$\frac{\sqrt[{\mathrm{4}}]{\mathrm{k}+\mathrm{1}}−\sqrt[{\mathrm{4}}]{\mathrm{k}}}{\left(\sqrt{\mathrm{k}+\mathrm{1}}+\sqrt{\mathrm{k}}\right)\left(\sqrt{\mathrm{k}+\mathrm{1}}−\sqrt{\mathrm{k}}\right)}\:=\:\sqrt[{\mathrm{4}}]{\mathrm{k}+\mathrm{1}}−\sqrt[{\mathrm{4}}]{\mathrm{k}} \\…
Question Number 103670 by bobhans last updated on 16/Jul/20 $${Given}\:{b}_{{n}} \:=\:\mathrm{3}.\mathrm{2}^{{n}} \:{is}\:{a}\:{GP}\:.\:{find}\:{the}\:{value} \\ $$$${of}\:\frac{\mathrm{1}}{{b}_{\mathrm{1}} }+\frac{\mathrm{1}}{{b}_{\mathrm{2}} }+\frac{\mathrm{1}}{{b}_{\mathrm{3}} }+…+\frac{\mathrm{1}}{{b}_{\mathrm{10}} }\:?\: \\ $$ Answered by bramlex last updated…
Question Number 103648 by Lordose last updated on 16/Jul/20 $$\boldsymbol{\mathrm{Evaluate}}\:\frac{\mathrm{1}}{\mathrm{1}\centerdot\mathrm{2}\centerdot\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{3}\centerdot\mathrm{4}\centerdot\mathrm{5}}+…+\frac{\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{1}}{\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right.} \\ $$ Answered by Dwaipayan Shikari last updated on 16/Jul/20 $${T}_{{n}} =\frac{\mathrm{2}{n}−\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$$\Sigma{T}_{{n}} =\Sigma\frac{{n}+{n}+\mathrm{2}−\mathrm{3}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\Sigma\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}+\Sigma\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}−\frac{\mathrm{3}}{\mathrm{2}}\Sigma\frac{\mathrm{2}+{n}−{n}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}…
Question Number 103633 by Lordose last updated on 16/Jul/20 $$ \\ $$$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{question}}\:\boldsymbol{\mathrm{is}} \\ $$$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{1}}{\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right.}\right)=… \\ $$ Commented by bobhans last updated on 16/Jul/20…
Question Number 38099 by Cheyboy last updated on 21/Jun/18 $${x}^{{x}} =\mathrm{0}.\mathrm{25} \\ $$$${find}\:{x} \\ $$ Commented by MrW3 last updated on 22/Jun/18 $${no}\:{real}\:{solution},\: \\ $$$${since}\:{x}^{{x}}…
Question Number 103623 by Lordose last updated on 16/Jul/20 $$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sum}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{series}}\:\boldsymbol{\mathrm{whose}}\:\boldsymbol{\mathrm{nth}} \\ $$$$\boldsymbol{\mathrm{term}}\:\boldsymbol{\mathrm{is}}\:\frac{\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{1}}{\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right.}. \\ $$$$\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{have}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{problem}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{need}} \\ $$$$\boldsymbol{\mathrm{help}}\:\boldsymbol{\mathrm{please}} \\ $$ Answered by OlafThorendsen last updated on 16/Jul/20…
Question Number 103620 by bobhans last updated on 16/Jul/20 $${If}\:{a}^{\mathrm{2}} −{bc},\:{b}^{\mathrm{2}} −{ac},\:{c}^{\mathrm{2}} −{ab}\:{is}\:{AP}\:{where}\:{a}+{c} \\ $$$$=\:\mathrm{12},\:{find}\:{the}\:{value}\:{of}\:{a}+{b}+{c}\: \\ $$ Answered by bemath last updated on 16/Jul/20 $$\Rightarrow{AP}\::\:\mathrm{2}\left({b}^{\mathrm{2}}…
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Question Number 103389 by M±th+et+s last updated on 14/Jul/20 Commented by M±th+et+s last updated on 14/Jul/20 $${find}\:{the}\:{hypotenuse}\:{of}\:{the}\:{triangle} \\ $$$${as}\:\left({x}\right) \\ $$ Answered by OlafThorendsen last…
Question Number 103321 by bemath last updated on 14/Jul/20 $$\mathrm{5}+\sqrt{\mathrm{5}−\sqrt{\mathrm{5}+\sqrt{\mathrm{5}−\sqrt{\mathrm{5}+\sqrt{\mathrm{5}−\sqrt{\mathrm{5}+…}}}}}} \\ $$ Answered by Dwaipayan Shikari last updated on 14/Jul/20 $$\sqrt{\mathrm{5}−\sqrt{\mathrm{5}+\sqrt{\mathrm{5}…}}}={p} \\ $$$$\mathrm{5}−\sqrt{\mathrm{5}+\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}…={p}^{\mathrm{2}} \\ $$$$\mathrm{5}+{p}=\left(\mathrm{5}−{p}^{\mathrm{2}}…