Question Number 130936 by mathlove last updated on 30/Jan/21 Answered by floor(10²Eta[1]) last updated on 30/Jan/21 $$\mathrm{15}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$$$\mathrm{so}\:\mathrm{15}^{\mathrm{n}} \equiv\mathrm{1}^{\mathrm{n}} =\mathrm{1}\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{3}^{\mathrm{2007}} } ×\mathrm{15}^{\mathrm{2}^{\mathrm{7777}}…
Question Number 130817 by shaker last updated on 29/Jan/21 Answered by Olaf last updated on 29/Jan/21 $$\mathrm{argth}\left(\mathrm{sh}^{\mathrm{4}} {x}+\mathrm{3sh}^{\mathrm{2}} {x}\right)\:=\:\mathrm{1} \\ $$$$\mathrm{sh}^{\mathrm{4}} {x}+\mathrm{3sh}^{\mathrm{2}} {x}\:=\:\mathrm{th}\left(\mathrm{1}\right)\:=\:\frac{{e}−\frac{\mathrm{1}}{{e}}}{{e}+\frac{\mathrm{1}}{{e}}}\:=\:\frac{{e}^{\mathrm{2}} −\mathrm{1}}{{e}^{\mathrm{2}} +\mathrm{1}}…
Question Number 65199 by rajesh4661kumar@gmail.com last updated on 26/Jul/19 Answered by Tanmay chaudhury last updated on 26/Jul/19 $$\left(\mathrm{1}\right),\left(\mathrm{2},\mathrm{3},\mathrm{4}\right),\left(\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right),\left(\mathrm{10},\mathrm{11},\mathrm{12},\mathrm{13},\mathrm{14},\mathrm{15},\mathrm{16}\right)… \\ $$$$\mathrm{1}{st}\:{group}\:\:{contains}\rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:{terms} \\ $$$$\mathrm{2}{nd}\:{group}\:{contains}\:\rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\:{terms} \\ $$$$\mathrm{3}{rd}\:\:\:{group}\:\:\:{contains}\rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}{terms} \\…
Question Number 65122 by aliesam last updated on 25/Jul/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 64955 by Tony Lin last updated on 23/Jul/19 $${prove}\:{that}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\centerdot\frac{\left({k}+\mathrm{1}\right)!}{\mathrm{2}^{{k}+\mathrm{1}} }=\frac{\left({n}+\mathrm{2}\right)!}{\mathrm{2}^{{n}+\mathrm{1}} }−\mathrm{1} \\ $$ Commented by ~ À ® @ 237 ~…
Question Number 64916 by Masumsiddiqui399@gmail.com last updated on 23/Jul/19 Commented by Tawa1 last updated on 23/Jul/19 $$\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right)^{\mathrm{x}} \:+\:\left(\frac{\mathrm{1}}{\mathrm{2}\:+\:\sqrt{\mathrm{3}}}\right)^{\mathrm{x}} \:=\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\mathrm{since}\:\:\:\:\mathrm{2}\:−\:\sqrt{\mathrm{3}}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}\:+\:\sqrt{\mathrm{3}}}\:\:\mathrm{on}\:\mathrm{rationalise}\right] \\ $$$$\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right)^{\mathrm{x}} \:+\:\frac{\mathrm{1}}{\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right)^{\mathrm{x}} }\:=\:\:\mathrm{4} \\ $$$$\mathrm{Let}\:\:\:\:\:\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right)^{\mathrm{x}}…
Question Number 64564 by Tawa1 last updated on 19/Jul/19 Commented by Tony Lin last updated on 19/Jul/19 $${let}\:{cosx}={t} \\ $$$$\frac{{t}}{\mathrm{1}+\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{{t}}}−\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\mathrm{1}+{t}} \\ $$$$=\frac{{t}^{\mathrm{2}} }{\mathrm{1}+\sqrt{\mathrm{1}−{t}^{\mathrm{2}}…
Question Number 64561 by aliesam last updated on 19/Jul/19 Commented by mathmax by abdo last updated on 20/Jul/19 $${let}\:{P}_{{n}} =\prod_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{1}+\sigma^{\mathrm{2}^{{k}} } \right)\:\Rightarrow{P}_{{n}} =\left(\mathrm{1}+\sigma\right)\left(\mathrm{1}+\sigma^{\mathrm{2}}…
Question Number 130085 by liberty last updated on 22/Jan/21 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{if}\:\begin{cases}{\mathrm{a}=\mathrm{r}^{\mathrm{2}} −\mathrm{2rs}−\mathrm{s}^{\mathrm{2}} }\\{\mathrm{b}=\mathrm{r}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} }\\{\mathrm{c}=\mathrm{r}^{\mathrm{2}} +\mathrm{2rs}−\mathrm{s}^{\mathrm{2}} }\end{cases} \\ $$$$\:\mathrm{for}\:\mathrm{some}\:\mathrm{integers}\:\mathrm{r},\mathrm{s}\:\mathrm{then}\:\mathrm{a}^{\mathrm{2}} ,\mathrm{b}^{\mathrm{2}} ,\mathrm{c}^{\mathrm{2}} \\ $$$$\mathrm{are}\:\mathrm{three}\:\mathrm{square}\:\mathrm{in}\:\mathrm{AP}. \\ $$ Answered…
Question Number 64465 by Theophilus111 last updated on 18/Jul/19 $${vy} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com