Question Number 108818 by bobhans last updated on 19/Aug/20 $$\:\:\frac{\boldsymbol{\mathcal{B}}{ob}\boldsymbol{{hans}}}{\Delta} \\ $$$$\left(\mathrm{1}\right){Let}\:{n}\:{be}\:{a}\:{positive}\:{integer},\:{and}\:{let}\:{x}\:{and}\:{y}\:{be}\:{positive}\:{real}\: \\ $$$${number}\:{such}\:{that}\:{x}^{{n}} \:+\:{y}^{{n}} \:=\:\mathrm{1}\:.\:{Prove}\:{that}\: \\ $$$$\left(\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}+{x}^{\mathrm{2}{k}} }{\mathrm{1}+{x}^{\mathrm{4}{k}} }\:\right)\left(\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}+{y}^{\mathrm{2}{k}} }{\mathrm{1}+{y}^{\mathrm{4}{k}}…
Question Number 108679 by Rasikh last updated on 18/Aug/20 Answered by Dwaipayan Shikari last updated on 18/Aug/20 $$\sqrt{\mathrm{1}+\mathrm{8}}\:=\mathrm{3} \\ $$$$\sqrt{\mathrm{1}+\mathrm{2}.\mathrm{4}}=\mathrm{3} \\ $$$$\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}.\mathrm{15}}}=\mathrm{3} \\ $$$$\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}}}….}}=\mathrm{3} \\…
Question Number 108594 by Rasikh last updated on 18/Aug/20 Answered by 1549442205PVT last updated on 18/Aug/20 $$\mathrm{From}\:\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0}\:\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}=\mathrm{1we}\:\mathrm{have} \\ $$$$\mathrm{ab}+\mathrm{bc}+\mathrm{ca}=\mathrm{abc}\left(\mathrm{1}\right).\mathrm{Apply}\:\mathrm{Cauchy}'\mathrm{s}\: \\ $$$$\mathrm{inequality}\:\mathrm{for}\:\mathrm{three}\:\mathrm{positive}\:\mathrm{numbers} \\ $$$$\mathrm{we}\:\mathrm{have}\:: \\ $$$$\mathrm{abc}=\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\geqslant\mathrm{3}\:^{\mathrm{3}}…
Question Number 43036 by Rio Michael last updated on 06/Sep/18 Commented by Rio Michael last updated on 06/Sep/18 $${Find}\:{the}\:{area}\:{of}\:{one}\:{square}\:{of}\:{the}\:{squares}\left(\frac{\mathrm{1}}{\mathrm{4}}\:{of}\:{the}\:{big}\:{square}\right) \\ $$ Answered by $@ty@m last…
Question Number 173902 by mnjuly1970 last updated on 20/Jul/22 $$ \\ $$$$\:\:{Q}:\:{How}\:{many}\:{common}\:{three}−{digit}\:{numbers} \\ $$$$\:\:\:\:{are}\:{there}\:{in}\:{the}\:{following} \\ $$$$\:\:\:\:\:{two}\:{sequences}? \\ $$$$\:\:\:\:\:\begin{cases}{\:\:{a}_{{n}} \:=\:\mathrm{1}\:\:,\:\mathrm{5}\:,\:\mathrm{9}\:,\mathrm{13}\:,\:…}\\{\:\:{b}_{\:{m}} \:=\:\mathrm{4}\:,\:\mathrm{7}\:,\:\mathrm{10}\:,\:\mathrm{13}\:,…}\end{cases} \\ $$ Answered by mr…
Question Number 173768 by BaliramKumar last updated on 17/Jul/22 $${Total}\:{factors}\:\:{of}\:\:\mathrm{20}!\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 42672 by Tawa1 last updated on 31/Aug/18 $$\mathrm{Simplify}:\:\:\:\left(\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\right)\left(\mathrm{x}^{−\mathrm{1}} \:+\:\mathrm{y}^{−\mathrm{1}} \:+\:\mathrm{z}^{−\mathrm{1}} \right)\:=\:\left(\mathrm{x}^{−\mathrm{1}} \:\mathrm{y}^{−\mathrm{1}} \:\mathrm{z}^{−\mathrm{1}} \right)\left(\mathrm{x}\:+\:\mathrm{y}\right)\left(\mathrm{y}\:+\:\mathrm{z}\right)\left(\mathrm{z}\:+\:\mathrm{x}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 108082 by ismoilov last updated on 14/Aug/20 Answered by Her_Majesty last updated on 14/Aug/20 $${x}={tan}^{−\mathrm{1}} {t} \\ $$$$\frac{\left({t}+\mathrm{1}\right)\left({t}−\mathrm{2}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}\geqslant\mathrm{0}\:\Rightarrow\:−\mathrm{1}\leqslant{t}\leqslant\mathrm{2} \\ $$$$\Rightarrow\:{for}\:\mathrm{0}\leqslant{x}<\mathrm{2}\pi \\ $$$$\frac{\pi}{\mathrm{2}}−{tan}^{−\mathrm{1}}…
Question Number 42495 by Tawa1 last updated on 26/Aug/18 $$\mathrm{In}\:\mathrm{the}\:\mathrm{sequence}\:\:\mathrm{1},\:\mathrm{22},\:\mathrm{333},\:…\:\mathrm{10101010101010101010},\:\mathrm{1111111111111111111111},\:… \\ $$$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{in}\:\mathrm{the}\:\mathrm{200th}\:\mathrm{term}\:\mathrm{is}\:?? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 107974 by Ar Brandon last updated on 13/Aug/20 $$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{k}\right)^{\mathrm{2}} }} \\ $$ Answered by Dwaipayan Shikari last updated on 14/Aug/20…