Question Number 152950 by john_santu last updated on 03/Sep/21 $$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\right)+\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{2}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}\right)+…+\left(\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{2}}{\mathrm{8}}+…+\frac{\mathrm{7}}{\mathrm{8}}\right)=? \\ $$ Commented by MJS_new last updated on 03/Sep/21 $$\mathrm{14} \\ $$ Terms of Service…
Question Number 152946 by john_santu last updated on 03/Sep/21 $$\:\:\:\frac{\mathrm{4038}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{15}}+…+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…+\mathrm{2019}}}\:=? \\ $$ Commented by mathdanisur last updated on 03/Sep/21 $$\blacktriangle\:\frac{\mathrm{2}\centerdot\mathrm{2019}}{\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}}\:+\:…\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:…\:+\:\mathrm{2019}}} \\ $$$$\boldsymbol{\mathrm{A}}\:=\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}}\:+\:…\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}\:+\:…\:+\:\mathrm{2019}} \\ $$$$\boldsymbol{\mathrm{A}}\:=\:\mathrm{1}\:+\:\frac{\mathrm{2}}{\mathrm{2}\centerdot\mathrm{3}}\:+\:\frac{\mathrm{2}}{\mathrm{3}\centerdot\mathrm{4}}\:+\:…\:+\:\frac{\mathrm{2}}{\mathrm{2019}\centerdot\mathrm{2020}} \\…
Question Number 21874 by tawa tawa last updated on 05/Oct/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{if}\:\:\:\mathrm{2}^{\mathrm{2006}} \:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\:\mathrm{17} \\ $$ Answered by mrW1 last updated on 06/Oct/17 $$\mathrm{2}^{\mathrm{2006}} =\mathrm{2}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{2004}} =\mathrm{4}×\left(\mathrm{2}^{\mathrm{4}}…
Question Number 21867 by Tinkutara last updated on 05/Oct/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{points}\:\mathrm{in}\:\mathrm{the}\:\mathrm{cartesian} \\ $$$$\mathrm{plane}\:\mathrm{with}\:\mathrm{integral}\:\mathrm{coordinates} \\ $$$$\mathrm{satisfying}\:\mathrm{the}\:\mathrm{inequalities}\:\mid{x}\mid\:\leqslant\:\mathrm{4},\:\mid{y}\mid\:\leqslant \\ $$$$\mathrm{4}\:\mathrm{and}\:\mid{x}\:−\:{y}\mid\:\leqslant\:\mathrm{4}\:\mathrm{is} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 87398 by jagoll last updated on 04/Apr/20 $$\mathrm{dear}\:\mathrm{mr}\:\mathrm{w} \\ $$$$\mathrm{a}_{\mathrm{n}+\mathrm{2}} \:=\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} \:−\:\mathrm{a}_{\mathrm{n}} \\ $$$$\mathrm{find}\:\mathrm{a}_{\mathrm{n}} \\ $$ Commented by mr W last updated on…
Question Number 87382 by mathocean1 last updated on 04/Apr/20 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{tan}\frac{\mathrm{5}\pi}{\mathrm{12}}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{this}\: \\ $$$$\mathrm{equation}:\:{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$ Answered by ajfour last updated on 04/Apr/20 $${x}^{\mathrm{3}} +\mathrm{1}=\mathrm{3}{x}\left({x}+\mathrm{1}\right)…
Question Number 87379 by john santu last updated on 04/Apr/20 $$\underset{\mathrm{k}\:=\:\mathrm{2}} {\overset{\mathrm{2010}} {\prod}}\:\frac{\mathrm{k}^{\mathrm{2}} −\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\:=\:? \\ $$ Commented by jagoll last updated on 04/Apr/20 $$\underset{\mathrm{k}\:=\:\mathrm{2}}…
Question Number 152912 by mathdanisur last updated on 03/Sep/21 Answered by ghimisi last updated on 03/Sep/21 $$\Sigma\frac{{a}}{\mathrm{2}{b}+\mathrm{3}{c}}=\Sigma\frac{{a}^{\mathrm{2}} }{\mathrm{2}{ab}+\mathrm{3}{ac}}\geqslant\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} }{\mathrm{5}\left({ab}+{bc}+{ac}\right)}\geqslant\frac{\mathrm{3}\left({ab}+{bc}+{ca}\right)}{\mathrm{5}\left({ab}+{bc}+{ac}\right)}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$ Commented by ghimisi last…
Question Number 152907 by mathdanisur last updated on 03/Sep/21 $$\mathrm{Find}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{form}: \\ $$$$\Omega=\left(\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{x}^{\mathrm{29}} −\mathrm{x}^{\mathrm{9}} }{\mathrm{x}^{\mathrm{40}} +\mathrm{1}}\:\mathrm{dx}\right)\left(\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{x}^{\mathrm{29}} −\mathrm{2x}^{\mathrm{9}} }{\mathrm{x}^{\mathrm{40}} +\mathrm{4}}\mathrm{dx}\right) \\ $$ Answered…
Question Number 152901 by ajfour last updated on 03/Sep/21 Terms of Service Privacy Policy Contact: info@tinkutara.com