Question Number 127486 by shaker last updated on 30/Dec/20 Answered by Dwaipayan Shikari last updated on 30/Dec/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{cos}\left({a}_{{k}} {x}\right)−\mathrm{1}}{{x}^{\mathrm{2}} }={y} \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 61922 by Tawa1 last updated on 11/Jun/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}:\:\:\:\:\:\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{1}}{\mathrm{n}\:+\:\mathrm{2}}.\:\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}!} \\ $$ Commented by maxmathsup by imad last updated on 12/Jun/19…
Question Number 127360 by bemath last updated on 29/Dec/20 Answered by liberty last updated on 29/Dec/20 $${let}\::\:{a}−\mathrm{7}{b},{a}−\mathrm{6}{b},{a}−\mathrm{5}{b},…,\:{a}+\mathrm{5}{b},\:{a}+\mathrm{6}{b},\:{a}+\mathrm{7}{b}\:{is}\:{AP} \\ $$$${given}\:{condition}\:\rightarrow\begin{cases}{\mathrm{3}{a}−\mathrm{18}{b}=−\mathrm{60};\:{a}−\mathrm{6}{b}=−\mathrm{20}\:\left({the}\:{first}\:\mathrm{3}\:{terms}\right)}\\{\mathrm{3}{a}+\mathrm{18}{b}=\mathrm{84};\:{a}+\mathrm{6}{b}\:=\:\mathrm{28}\:\left({the}\:{last}\:\mathrm{3}\:{terms}\right)}\end{cases} \\ $$$${we}\:{get}\:\begin{cases}{{a}=\mathrm{4}}\\{{b}=\mathrm{4}}\end{cases}.\:{we}\:{want}\:{to}\:{compute}\:{the} \\ $$$${sum}\:{of}\:{the}\:{middle}\:\mathrm{3}\:{terms}\:\Rightarrow{T}_{\mathrm{7}} +{T}_{\mathrm{8}} +{T}_{\mathrm{9}}…
Question Number 192893 by mnjuly1970 last updated on 30/May/23 $$ \\ $$$$\:\:\:\:\:\:\mathrm{Q}\::\:\mathrm{Find}\:\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{of}\:\:\mathrm{dividing} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{the}\:\mathrm{following}\:\mathrm{number}\:\mathrm{by}\:\:\mathrm{7}\:. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{N}\:=\:\mathrm{3}^{\:\mathrm{10}^{\:\mathrm{1}} } \:+\:\mathrm{3}^{\:\mathrm{10}^{\:\mathrm{2}} \:} \:+\:\mathrm{3}^{\:\mathrm{10}^{\:\mathrm{3}} \:} \:+\:…\:+\:\mathrm{3}^{\:\mathrm{10}^{\:\mathrm{10}} }…
Question Number 61692 by Tony Lin last updated on 06/Jun/19 $${if}\frac{\mathrm{3}}{\mathrm{2}}\leqslant{x}\leqslant\mathrm{5}, \\ $$$${prove}:\mathrm{2}\sqrt{{x}+\mathrm{1}}+\sqrt{\mathrm{2}{x}−\mathrm{3}}+\sqrt{\mathrm{15}−\mathrm{3}{x}}<\mathrm{2}\sqrt{\mathrm{19}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 61479 by Tawa1 last updated on 03/Jun/19 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}:\:\:\:\:\:\:\:\frac{\mathrm{6}\sqrt{\mathrm{2x}}}{\mathrm{x}\:−\:\mathrm{1}}\:+\:\frac{\mathrm{5}\sqrt{\mathrm{x}\:−\:\mathrm{1}}}{\mathrm{2x}}\:\:\:=\:\:\mathrm{13} \\ $$ Commented by MJS last updated on 03/Jun/19 $$\mathrm{we}\:\mathrm{cannot}\:\mathrm{generally}\:\mathrm{solve}\:\mathrm{this}… \\ $$ Answered by ajfour…
Question Number 192420 by Mingma last updated on 17/May/23 Commented by AST last updated on 18/May/23 $${Q}\mathrm{191758} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 61313 by Tony Lin last updated on 31/May/19 $$\left.{how}\:{to}\:{calculate}\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}0}.\mathrm{05}^{\mathrm{0}.\mathrm{05}^{\mathrm{0}.\mathrm{05}^{.^{.^{.\mathrm{0}.\mathrm{05}} } } } } \right\}{n} \\ $$$${in}\:{a}\:{simple}\:{and}\:{fast}\:{way}? \\ $$ Commented by mr W…
Question Number 192367 by Rupesh123 last updated on 15/May/23 Answered by mehdee42 last updated on 15/May/23 $${t}_{{r}} ={r}−\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow{t}_{{r}+\mathrm{1}} ={r}−\frac{\mathrm{1}}{\mathrm{2}}\:\&\:{t}_{{r}+\mathrm{2}} ={r}+\frac{\mathrm{1}}{\mathrm{2}}\&\:{t}_{{r}+\mathrm{3}} ={r}+\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{t}_{{r}} {t}_{{r}+\mathrm{1}} {t}_{{r}+\mathrm{2}}…