Question Number 125640 by aurpeyz last updated on 12/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 125641 by aurpeyz last updated on 12/Dec/20 Answered by Dwaipayan Shikari last updated on 12/Dec/20 $${S}=\mathrm{2}+\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{8}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{11}}{\mathrm{2}^{\mathrm{3}} }+… \\ $$$$\frac{{S}}{\mathrm{2}}=\:\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{8}}{\mathrm{2}^{\mathrm{3}} }+.. \\…
Question Number 125638 by aurpeyz last updated on 12/Dec/20 Answered by mathmax by abdo last updated on 12/Dec/20 $$\mathrm{S}\:=\frac{\mathrm{3}}{\mathrm{3}}−\frac{\mathrm{5}}{\mathrm{5}^{\mathrm{2}} }+\frac{\mathrm{7}}{\mathrm{5}^{\mathrm{3}} }−\frac{\mathrm{9}}{\mathrm{5}^{\mathrm{4}} }+…\Rightarrow\mathrm{S}\:=\mathrm{1}+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{a}_{\mathrm{n}}…
Question Number 125639 by aurpeyz last updated on 12/Dec/20 Commented by ZiYangLee last updated on 13/Dec/20 $$\mathrm{A}\:\mathrm{more}\:\mathrm{alternative}\:\mathrm{ways}: \\ $$$$\:\:\:\:\:{f}\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{{n}} =\frac{\mathrm{1}}{\mathrm{1}−{x}}−\mathrm{1} \\ $$$$\Rightarrow{f}'\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty}…
Question Number 125636 by Eric002 last updated on 12/Dec/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{{sin}\:{x}} {sin}\frac{\mathrm{1}}{{t}}\:{cos}\:{t}^{\mathrm{2}} \:{dt} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 125637 by aurpeyz last updated on 12/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191169 by Mingma last updated on 19/Apr/23 Answered by mehdee42 last updated on 19/Apr/23 $${c}:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}\:\Rightarrow{m}_{{l}} ={y}'_{{A}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\Rightarrow\:{l}:\:\:{y}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left({x}−\mathrm{4}\right)\: \\ $$$${m}_{{d}} ×{m}_{{l}} =−\mathrm{1}\Rightarrow{m}_{{l}}…
Question Number 191168 by Mingma last updated on 19/Apr/23 Answered by mehdee42 last updated on 19/Apr/23 $${A}={lim}_{{n}\rightarrow\infty} \frac{\sqrt[{{n}}]{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)…\left(\mathrm{2}{n}\right)}}{{n}}={lim}_{{n}\rightarrow\infty} \sqrt[{{n}}]{\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)…\left(\mathrm{2}{n}\right)}{{n}^{{n}} }} \\ $$$$\Rightarrow{lnA}={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}\left[{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)+{ln}\left(\mathrm{1}+\frac{\mathrm{2}}{{n}}\right)+…+{ln}\left(\mathrm{1}+\frac{{n}}{{n}}\right)\right] \\ $$$$={lim}_{{n}\rightarrow\infty}…
Question Number 125632 by Mathgreat last updated on 12/Dec/20 Answered by snipers237 last updated on 13/Dec/20 $${CosA}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}\:\Rightarrow\:{sin}^{\mathrm{2}} \left(\frac{{A}}{\mathrm{2}}\right)=\frac{\mathrm{1}−{cosA}}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} }{\mathrm{4}{bc}}<\frac{{a}^{\mathrm{2}} }{\mathrm{4}{bc}} \\…
Question Number 191165 by Mingma last updated on 19/Apr/23 Terms of Service Privacy Policy Contact: info@tinkutara.com