Question Number 217190 by efronzo1 last updated on 05/Mar/25 $$\mathrm{Given}\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} \:=\:\mathrm{a}_{\mathrm{n}} \:+\:\mathrm{a}_{\mathrm{n}+\mathrm{2}} \: \\ $$$$\:\:\mathrm{where}\:\mathrm{a}_{\mathrm{3}} =\:\mathrm{4}\:\mathrm{and}\:\mathrm{a}_{\mathrm{5}} =\:\mathrm{6} \\ $$$$\:\mathrm{find}\:\mathrm{a}_{\mathrm{n}} \:. \\ $$ Answered by mathmax…
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Question Number 216953 by mustaphapelumi last updated on 25/Feb/25 Commented by Ghisom last updated on 26/Feb/25 $$\mathrm{or}\:\mathrm{use}\:\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}}{{x}}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}\left[\mathrm{sin}\:{x}\right]}{{dx}}}{\frac{{d}\left[{x}\right]}{{dx}}}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}}{\mathrm{1}}\:=\mathrm{1} \\ $$…
Question Number 216926 by Engr_Jidda last updated on 24/Feb/25 $${Evaluate}\:\mathrm{5}^{\mathrm{2}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{{m}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{m}^{\mathrm{2}} +\mathrm{2}{m}}\right)^{{n}−\mathrm{1}} \\ $$ Answered by Wuji last updated on 25/Feb/25…
Question Number 216925 by Engr_Jidda last updated on 24/Feb/25 Answered by mehdee7396 last updated on 24/Feb/25 $${s}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{4}{u}+\mathrm{1}\right){du}+\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{4}{u}+\mathrm{1}\right){du}+…+\int_{\mathrm{0}} ^{\mathrm{10}} \left(\mathrm{4}{u}+\mathrm{1}\right){du} \\ $$$$\left.=\left.\left(\left.\mathrm{2}{u}^{\mathrm{2}}…
Question Number 216917 by alcohol last updated on 24/Feb/25 $${li}\underset{{x}\rightarrow+\infty} {{m}}\:\left(\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}}−\sqrt{{x}}\right) \\ $$ Answered by mehdee7396 last updated on 24/Feb/25 $${lim}_{{x}\rightarrow\infty} \:\frac{\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}}{\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}+\sqrt{{x}}}} \\ $$$$={lim}_{{x}\rightarrow\infty} \:\frac{\sqrt{{x}}}{\:\mathrm{2}\sqrt{{x}}}=\frac{\mathrm{1}}{\mathrm{2}}…
Question Number 216748 by sniper237 last updated on 17/Feb/25 $${Find}\:\:\int_{−\mathrm{1}} ^{\mathrm{1}} {ln}\mid\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+{it}\right)\mid{dt} \\ $$ Answered by issac last updated on 18/Feb/25 $$\mathrm{can}'\mathrm{t}\:\mathrm{simplify} \\ $$$$\mathrm{function}\:\mathrm{ln}\left(\mid\Gamma\left(\boldsymbol{{i}}{t}+\frac{\mathrm{1}}{\mathrm{2}}\right)\mid\right) \\…
Question Number 216274 by mnjuly1970 last updated on 02/Feb/25 $$ \\ $$$$\:\:\:{prove}\:{that}\:: \\ $$$$ \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:{cos}\left(\:{n}\:\right)}{{n}}\:\left(\:\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\:…+\frac{\mathrm{1}}{\:\sqrt{{n}}}\:\right) \\ $$$$\:\: \\ $$$$\:\:\:\:\:\:{is}\:\:\:{convergent}. \\ $$$$ \\…
Question Number 216055 by efronzo1 last updated on 26/Jan/25 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x}}{\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\mathrm{x}}−\sqrt[{\mathrm{4}}]{\mathrm{cos}\:\mathrm{x}}}\:=? \\ $$ Answered by golsendro last updated on 27/Jan/25 $$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)}{\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\mathrm{x}}−\sqrt[{\mathrm{4}}]{\mathrm{cos}\:\mathrm{x}}}…
Question Number 216032 by efronzo1 last updated on 26/Jan/25 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}\:\sqrt{\mathrm{cos}\:\mathrm{2x}}}{\mathrm{x}^{\mathrm{2}} }\:=? \\ $$ Answered by mr W last updated on 26/Jan/25 $$\mathrm{cos}\:{x}\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\…