Question Number 24359 by math solver last updated on 16/Nov/17 Commented by ajfour last updated on 17/Nov/17 $${If}\:\underset{{r}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{9}{r}^{\mathrm{2}} +\mathrm{3}{r}−\mathrm{1}}\right) \\ $$$$\:\:\:\:=\mathrm{cot}^{−\mathrm{1}} \left(\frac{{m}}{{n}}\right)\:,\:{then}\:{it}\:{means}…
Question Number 155420 by amin96 last updated on 30/Sep/21 $$\frac{\mathrm{1}}{\mathrm{42}}+\frac{\mathrm{1}}{\mathrm{72}}+\frac{\mathrm{1}}{\mathrm{110}}+\ldots=? \\ $$ Answered by puissant last updated on 01/Oct/21 $${S}=\frac{\mathrm{1}}{\mathrm{42}}+\frac{\mathrm{1}}{\mathrm{72}}+\frac{\mathrm{1}}{\mathrm{110}}+…….=\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{9}}+…. \\ $$$$\Rightarrow\:{S}=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{2}} {\overset{\infty}…
Question Number 155393 by cortano last updated on 30/Sep/21 $$\:\:\mid\mid\mathrm{x}−\mathrm{4}\mid−\mathrm{4}\mid\:\geqslant\:\mathrm{4}\: \\ $$ Answered by mr W last updated on 30/Sep/21 $$\mid{x}−\mathrm{4}\mid−\mathrm{4}\geqslant\mathrm{4}\: \\ $$$$\Rightarrow\mid{x}−\mathrm{4}\mid\geqslant\mathrm{8}\: \\ $$$$\Rightarrow{x}−\mathrm{4}\geqslant\mathrm{8}\:{or}\:{x}−\mathrm{4}\leqslant−\mathrm{8}\:…
Question Number 155386 by mathdanisur last updated on 29/Sep/21 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{H}_{\boldsymbol{\mathrm{k}}} \right)^{\mathrm{3}} }{\mathrm{2}^{\boldsymbol{\mathrm{k}}} }\:=\:\frac{\mathrm{3}\zeta\left(\mathrm{3}\right)\:+\:\mathrm{ln}^{\mathrm{3}} \mathrm{2}\:+\:\pi^{\mathrm{2}} \mathrm{ln2}}{\mathrm{3}}\:\: \\ $$ Answered by Kamel last…
Question Number 155377 by mathdanisur last updated on 29/Sep/21 $$\mathrm{For}\:\mathrm{two}\:\mathrm{non}-\mathrm{negative}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{a}^{\mathrm{6}} \:+\:\mathrm{b}^{\mathrm{6}} \:=\:\mathrm{2} \\ $$$$\mathrm{Find}\:\:\mathrm{a};\mathrm{b}\:\:\mathrm{such}\:\mathrm{that}\:\:\mathrm{3}\left(\mathrm{a}+\mathrm{b}\right)=\mathrm{1}+\mathrm{5}\sqrt{\mathrm{ab}} \\ $$ Answered by MJS_new last updated on 30/Sep/21…
Question Number 155372 by mathdanisur last updated on 29/Sep/21 Commented by Rasheed.Sindhi last updated on 30/Sep/21 $${What}'{s}\:{meant}\:{by}\:\mathrm{Z}.\mathrm{Q}\:? \\ $$ Commented by mathdanisur last updated on…
Question Number 89834 by Cheyboy last updated on 19/Apr/20 $${Find}\:{x}\: \\ $$$$\boldsymbol{{e}}^{\boldsymbol{{x}}} =\:\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{1} \\ $$$$\boldsymbol{{anyother}}\:\boldsymbol{{method}}\:\boldsymbol{{apart}}\:\boldsymbol{{from}}\:\boldsymbol{{N}}{ewton}'{s} \\ $$ Commented by Joel578 last updated on 19/Apr/20…
Question Number 89829 by student work last updated on 19/Apr/20 Commented by MJS last updated on 19/Apr/20 English please / Français s'il vous plaît Commented by student work last updated on…
Question Number 89826 by I want to learn more last updated on 19/Apr/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} \:\:+\:\:\mathrm{xy}\:\:+\:\:\mathrm{y}^{\mathrm{2}} \:\:\:=\:\:\:\mathrm{7}\:\:\:\:\:\:\:\:\:……\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\mathrm{y}^{\mathrm{2}} \:\:+\:\:\mathrm{yz}\:\:+\:\:\mathrm{z}^{\mathrm{2}} \:\:\:=\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:……\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\:\:\mathrm{z}^{\mathrm{2}} \:\:+\:\:\mathrm{xz}\:\:+\:\:\mathrm{x}^{\mathrm{2}}…
Question Number 155362 by mathdanisur last updated on 29/Sep/21 $$\mathrm{if}\:\:\mathrm{a};\mathrm{b}\geqslant\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{a}^{\mathrm{4}} \:+\:\mathrm{b}^{\mathrm{4}} \:=\:\mathrm{17}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{15}\left(\mathrm{a}\:+\:\mathrm{b}\right)\:\geqslant\:\mathrm{17}\:+\:\mathrm{14}\sqrt{\mathrm{2ab}} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com