Question Number 99011 by bemath last updated on 18/Jun/20 Commented by MJS last updated on 18/Jun/20 $$\mathrm{strange}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{question} \\ $$$$\mathrm{the}\:\mathrm{given}\:\mathrm{answers}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3}\:\mathrm{are}\:\mathrm{one}\:\mathrm{possible} \\ $$$$\mathrm{solution}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{an} \\ $$$$\mathrm{unique}\:\mathrm{solution} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{\mathrm{16}}…
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Question Number 98848 by M±th+et+s last updated on 16/Jun/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+{k}+\mathrm{1}} \left(\mathrm{1}+{k}\right)^{\mathrm{2}} }{{n}\left({n}+{k}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 164189 by alf123 last updated on 15/Jan/22 Commented by MJS_new last updated on 15/Jan/22 $$\mathrm{no}\:\mathrm{nice}\:\mathrm{solution}.\:\mathrm{you}'\mathrm{ll}\:\mathrm{have}\:\mathrm{to}\:\mathrm{approximate} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 33088 by Joel578 last updated on 10/Apr/18 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{n}^{\mathrm{2}} }{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$ Commented by abdo imad last updated on…
Question Number 98557 by M±th+et+s last updated on 14/Jun/20 $${a}_{{n}} =\underset{{k}=\mathrm{1}\:} {\overset{{n}−\mathrm{1}} {\sum}}\frac{{sin}\left(\frac{\left(\mathrm{2}{k}−\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)}{{cos}^{\mathrm{2}} \left(\frac{\left({k}−\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right){cos}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}}\right)} \\ $$$$ \\ $$$${find}\:\:\underset{{n}\rightarrow\infty} {{lim}}\frac{{a}_{{n}} }{{n}^{\mathrm{3}} } \\ $$ Terms…
Question Number 98434 by M±th+et+s last updated on 14/Jun/20 $${prove}\:{that} \\ $$$$\Omega=\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\underset{{m}=\mathrm{0}\:} {\overset{+\infty} {\sum}}\frac{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right).\Gamma\left({m}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}+\mathrm{1}\right).\Gamma\left({m}+\mathrm{1}\right)}.\frac{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}} .\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{m}} }{\left({n}+{m}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{2}}}.{G}_{\mathrm{2},\mathrm{2}} ^{\mathrm{2},\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\mid_{\mathrm{0}\:\:,\:\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}} \right) \\…
Question Number 98342 by Lekhraj last updated on 13/Jun/20 Commented by MJS last updated on 13/Jun/20 $$\mathrm{29} \\ $$$${n}=\mathrm{29}+\mathrm{30}{k} \\ $$ Commented by Lekhraj last…
Question Number 98309 by Rio Michael last updated on 12/Jun/20 $$\mathrm{let}\:{f}\left({x}\right)\:=\:\mathrm{ln}\left(\:\mathrm{1}\:+\:\mathrm{cosh}\:\mathrm{2}{x}\right) \\ $$$$\mathrm{show}\:\mathrm{that}\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{f}\left({x}\right)\:=\:\mathrm{2}{x}\:−\:\mathrm{ln}\:\mathrm{2} \\ $$$$\mathrm{hence}\:\mathrm{deduce}\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{f}\left({x}\right)\:\mathrm{with}\:\mathrm{the}\:\mathrm{asympotes}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{curve}. \\ $$ Answered by abdomathmax last…
Question Number 98286 by M±th+et+s last updated on 12/Jun/20 $${justify}: \\ $$$$\underset{{x}\rightarrow+\infty} {{lim}}\:\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left({k}−\mathrm{1}\right)!\:{sin}\left({x}−\frac{\pi}{\mathrm{2}}{k}\right)}{{x}^{{k}} }=\frac{\pi}{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com