Question Number 195904 by York12 last updated on 13/Aug/23 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{2}^{{n}} \left(\mathrm{2}{n}\right)!}{\mathrm{3}^{\mathrm{2}{n}+\mathrm{1}} \left({n}+\mathrm{1}\right)!{n}!}\right]=\lambda \\ $$$${Evaluate}\:\left(\lambda\right) \\ $$ Answered by qaz last updated on 13/Aug/23…
Question Number 195898 by Calculusboy last updated on 12/Aug/23 Answered by qaz last updated on 13/Aug/23 $${a}_{{n}+\mathrm{2}} {a}_{{n}+\mathrm{1}} −{a}_{{n}+\mathrm{1}} {a}_{{n}} =\mathrm{2}\:\:\:\:\Rightarrow{a}_{{n}+\mathrm{2}} {a}_{{n}+\mathrm{1}} =\mathrm{2}{n}+{a}_{\mathrm{1}} {a}_{\mathrm{2}} =\mathrm{2}\left({n}+\mathrm{1}\right)…
Question Number 195820 by York12 last updated on 11/Aug/23 $${a},{b},{c}>\mathrm{0}\:\&{abc}=\mathrm{1},{prove}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{a}+{b}}+\frac{\mathrm{1}}{\mathrm{1}+{b}+{c}}+\frac{\mathrm{1}}{\mathrm{1}+{c}+{a}}\leqslant\mathrm{1} \\ $$ Answered by CrispyXYZ last updated on 12/Aug/23 $$\mathrm{let}\:{a}={x}^{\mathrm{3}} ,\:{b}={y}^{\mathrm{3}} ,\:{c}={z}^{\mathrm{3}} ,\:\mathrm{then}\:{xyz}=\mathrm{1}.…
Question Number 195855 by jabarsing last updated on 11/Aug/23 $$\begin{cases}{\mathrm{3}\sqrt{\sqrt[{\mathrm{3}}]{\mathrm{12}}−\sqrt[{\mathrm{3}}]{\mathrm{3}}}=\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{3}}]{{y}}−\sqrt[{\mathrm{3}}]{{z}}}\\{{x},{y},{z}\:\in\:{N}}\end{cases}\:\:\Rightarrow\:{x},{y},{z}\:=? \\ $$$${please}\:{help}\:{me} \\ $$ Answered by Rasheed.Sindhi last updated on 13/Aug/23 $$\mathrm{Unsuccessful}\:\mathrm{Try}… \\ $$$$\begin{cases}{\mathrm{3}\sqrt{\sqrt[{\mathrm{3}}]{\mathrm{12}}−\sqrt[{\mathrm{3}}]{\mathrm{3}}}=\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{3}}]{{y}}−\sqrt[{\mathrm{3}}]{{z}}}\\{{x},{y},{z}\:\in\:{N}}\end{cases}\:\:\Rightarrow\:{x},{y},{z}\:=? \\…
Question Number 195813 by jabarsing last updated on 11/Aug/23 $$\begin{cases}{\mathrm{3}\sqrt{\sqrt[{\mathrm{3}}]{\mathrm{12}}−\sqrt[{\mathrm{3}}]{\mathrm{3}}}\:\:=\:\sqrt[{\mathrm{3}}]{{x}}\:+\:\sqrt[{\mathrm{3}}]{{y}}\:−\sqrt[{\mathrm{3}}]{{z}}}\\{{x},{y},{z}\:\in\:{N}}\end{cases}\:\Rightarrow\:{x},{y},{z}\:=? \\ $$$${mr}.{W}\:{please}\:{help}\:{me} \\ $$$${and}\:{other}\:{my}\:{friends}\:{please}\:{help}\:{me} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195809 by mr W last updated on 11/Aug/23 $${if}\:{x}^{\mathrm{5}} +{x}+\mathrm{1}=\mathrm{0},\:{find}\:{x}^{\mathrm{3}} −{x}^{\mathrm{2}} =? \\ $$ Commented by jabarsing last updated on 11/Aug/23 $${hello}\:{dear}\:{W}, \\…
Question Number 195765 by SajaRashki last updated on 10/Aug/23 $${hello} \\ $$$$ \\ $$$$\:\begin{cases}{{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:\mathrm{18}}\\{{x}>\mathrm{1}}\end{cases}\:\:\Rightarrow\:\:\:{x}^{\mathrm{5}} −\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\:=\:? \\ $$ Answered by mr W last…
Question Number 195733 by York12 last updated on 09/Aug/23 $${prove}\:{that} \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left[\frac{{B}_{\overset{\_} {{n}}} }{\left({n}−\mathrm{2}\right)!}\right]=\frac{{e}\left(\mathrm{3}−{e}\right)}{\left({e}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${where}\:{B}_{\overset{\_} {{n}}} \:{is}\:{the}\:{n}−\:{th}\:{bernouli}'{s}\:{number} \\ $$ Answered by…
Question Number 195693 by SajaRashki last updated on 08/Aug/23 $${hello} \\ $$$$\left[\underset{{n}=\mathrm{1}} {\overset{\mathrm{10000}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{{n}}}\right]=? \\ $$$$\left[\:\right]\::\:{is}\:{bracket} \\ $$$${thank}\:{you} \\ $$$$ \\ $$ Commented by York12…
Question Number 195689 by Shomurotovdiyorbek last updated on 07/Aug/23 $$\int_{\mathrm{1}} ^{\mathrm{3}} {f}\left({x}\right)^{\mathrm{3}} {f}'\left({x}\right){dx}=\int_{\mathrm{1}} ^{\mathrm{3}} \left({f}\left({x}\right)^{\mathrm{3}} \right){df}\left({x}\right)=\int_{\mathrm{1}} ^{\mathrm{3}} {t}^{\mathrm{3}} {dt}=\frac{{t}^{\mathrm{4}} }{\mathrm{4}}\int_{\mathrm{1}} ^{\mathrm{3}} =\frac{{f}\left({x}\right)^{\mathrm{4}} }{\mathrm{4}}\int_{\mathrm{1}} \mathrm{3} \\…