Question Number 120978 by ZiYangLee last updated on 04/Nov/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\pi}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{2}\pi}+\ldots+\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}\pi}\right)=? \\ $$ Commented by Dwaipayan Shikari last updated on 04/Nov/20 $${If}\:{the}\:{question}\:{becomes}…
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Question Number 120972 by 0963808481 last updated on 04/Nov/20 $$\mathrm{2}^{{x}+\mathrm{1}} =\mathrm{8} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 186500 by 073 last updated on 05/Feb/23 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}}−\sqrt{\mathrm{x}}=? \\ $$$$\mathrm{please}\:\mathrm{solution} \\ $$ Answered by greougoury555 last updated on 05/Feb/23 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}}{\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}}\:+\sqrt{{x}}} \\…
Question Number 120952 by abony1303 last updated on 04/Nov/20 $$\mathrm{x},\mathrm{y},\mathrm{z}\:{are}\:{real}\:{numbers} \\ $$$$\begin{cases}{\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{4}}\\{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{10}}\\{\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} +\mathrm{z}^{\mathrm{3}} =\mathrm{22}}\end{cases}\:\:\:\:\:{Find}\:\mathrm{x}^{\mathrm{4}} +\mathrm{y}^{\mathrm{4}} +\mathrm{z}^{\mathrm{4}} \\ $$ Answered by MJS_new…
Question Number 55411 by naka3546 last updated on 23/Feb/19 Answered by kaivan.ahmadi last updated on 25/Feb/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 55410 by naka3546 last updated on 23/Feb/19 Answered by mr W last updated on 24/Feb/19 $${n}\left({n}−\mathrm{2016}\right)={m}^{\mathrm{2}} \\ $$$${n}^{\mathrm{2}} −\mathrm{2016}{n}−{m}^{\mathrm{2}} =\mathrm{0} \\ $$$${n}=\mathrm{1008}+\sqrt{\mathrm{1008}^{\mathrm{2}} +{m}^{\mathrm{2}}…
Question Number 120937 by Khalmohmmad last updated on 04/Nov/20 Answered by Dwaipayan Shikari last updated on 04/Nov/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}+\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty}…
Question Number 55401 by Kunal12588 last updated on 23/Feb/19 $${any}\:{wanna}\:{blow}\:{their}\:{mind} \\ $$ Commented by Kunal12588 last updated on 23/Feb/19 https://m.youtube.com/watch?v=OkmNXy7er84 Commented by Kunal12588 last updated…
Question Number 186468 by aba last updated on 04/Feb/23 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Simplify} \\ $$$$\frac{\mathrm{1}^{\mathrm{2}} \centerdot\mathrm{2}!+\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{3}!+\mathrm{3}^{\mathrm{2}} \centerdot\mathrm{4}!+\centerdot\centerdot\centerdot+{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)!−\mathrm{2}}{\left({n}+\mathrm{1}\right)!} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{to} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{n}^{\mathrm{2}} +\mathrm{n}−\mathrm{2} \\ $$ Answered by…