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Question-124534

Question Number 124534 by Novitasari last updated on 04/Dec/20 Commented by bemath last updated on 04/Dec/20 $$\mathrm{5}\sqrt{\mathrm{3}}+\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{3}}\:=\:\mathrm{5}\sqrt{\mathrm{3}} \\ $$ Terms of Service Privacy Policy Contact:…

Question-190067

Question Number 190067 by DAVONG last updated on 26/Mar/23 Answered by mehdee42 last updated on 26/Mar/23 $${S}=\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+…+\left[\sqrt{\mathrm{500}}\right] \\ $$$$\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]=\mathrm{3}×\mathrm{1} \\ $$$$\left[\sqrt{\mathrm{4}}\right]+\left[\sqrt{\mathrm{5}}\right]+\left[\mathrm{6}\right]+\left[\sqrt{\mathrm{7}}\right]+\left[\sqrt{\mathrm{8}}\right]=\mathrm{5}×\mathrm{2} \\ $$$$\vdots \\ $$$$\left[\sqrt{\mathrm{441}}\right]+\left[\sqrt{\mathrm{442}}\right]+…+\left[\sqrt{\mathrm{483}}\right]=\mathrm{43}×\mathrm{21}…

Question-190061

Question Number 190061 by sonukgindia last updated on 26/Mar/23 Answered by a.lgnaoui last updated on 29/Mar/23 $$\mathrm{2}\left({n}^{{n}+\mathrm{1}} −{n}!\right)=\mathrm{5}{n}^{\mathrm{2}} −\mathrm{3}{n}−\mathrm{2} \\ $$$$\frac{\mathrm{2}\left({n}^{{n}+\mathrm{1}} \right)}{{n}!}−\mathrm{1}=\frac{{a}}{{n}!}\left({a}=\right) \\ $$$$\frac{\mathrm{2}{n}×{n}^{{n}} }{{n}!}=\frac{\mathrm{5}\left({n}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}}…

Question-190047

Question Number 190047 by MathsFan last updated on 26/Mar/23 Commented by MathsFan last updated on 26/Mar/23 $$\:\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{find}\:\mathrm{the}\:\mathrm{forces}\: \\ $$$$\:\mathrm{in}\:\mathrm{all}\:\mathrm{members} \\ $$ Terms of Service Privacy…

a-b-c-1-a-b-c-1-Prove-that-1-a-2-1-1-b-2-1-1-c-2-1-27-10-

Question Number 58964 by naka3546 last updated on 02/May/19 $${a}\:+\:{b}\:+\:{c}\:\:=\:\:\mathrm{1} \\ $$$${a},\:{b},\:{c}\:\:\leqslant\:\:\mathrm{1} \\ $$$${Prove}\:\:{that} \\ $$$$\:\:\:\:\:\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} \:+\:\mathrm{1}}\:\:+\:\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} \:+\:\mathrm{1}}\:\:+\:\:\frac{\mathrm{1}}{{c}^{\mathrm{2}} \:+\:\mathrm{1}}\:\:\leqslant\:\:\frac{\mathrm{27}}{\mathrm{10}} \\ $$ Answered by tanmay last…