Category: Algebra

Question Number 205885 by hardmath last updated on 01/Apr/24 $$\mathrm{Find}:\underset{\mathbf{n}\to \mathrm{\infty }}{\lim }\sqrt[\mathbf{n}]{}\left(\begin{array}{c}2\mathrm{n}\\ \mathrm{n}\end{array}\right)=?$$ Answered by MM42 last updated on 03/Apr/24 $$\left(\begin{array}{c}2n\\ n\end{array}\right)=\frac{\left(2n\right)(2n-1)(2n-2)\dots (n+3)(n+2)(n+1)}{n(n-1)(n-2)\dots 3\times 2\times 1}$$$$a=\sqrt[n]{}\left(\begin{array}{c}2n\\ n\end{array}\right)$$$$\Rightarrow{lna}=\frac{\mathrm{1}}{{n}}\left[{ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{2}+\frac{\mathrm{1}}{{n}−\mathrm{1}}\right)+{ln}\left(\mathrm{2}+\frac{\mathrm{2}}{{n}−\mathrm{2}}\right)+…+{ln}\left(\mathrm{2}+\frac{{n}−\mathrm{3}}{\mathrm{3}}\right)\left(\mathrm{2}+\frac{{n}−\mathrm{2}}{\mathrm{2}}\right)\left(\mathrm{2}+\frac{{n}−\mathrm{1}}{\mathrm{1}}\right)\right.…

Question Number 205772 by mr W last updated on 30/Mar/24 Answered by MM42 last updated on 30/Mar/24 $$S_n=(\sqrt{1\times 2}-1)$$$$+(\sqrt{2\times 3}-\sqrt{1\times 2}-1)$$$$+(\sqrt{3\times 4}-\sqrt{2\times 3}-1)$$$$\vdots…

Question Number 205789 by hardmath last updated on 30/Mar/24 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 205767 by lmcp1203 last updated on 30/Mar/24 $$$$$$a,b,c\in {\mathrm{\Re }}^{+}$$$$a+b+c=1$$$$a^2/(1+b+c)+b^2/(1+a+c)+c^2/(1+a+b)⩾k$$$$findkmax.$$$${hint}\::\:{inequality}\:{cauchy}\:{schwarz} \…

Question Number 205746 by Satyam1234 last updated on 29/Mar/24 Answered by A5T last updated on 29/Mar/24 $$C\ne J+2\Rightarrow C=3F$$$$A⩾1;A=1\Rightarrow B=3\Rightarrow I=7\Rightarrow H=4\Rightarrow A=8X$$$$A=2\Rightarrow B=6\Rightarrow E=2X$$$$A=3\Rightarrow B=9X(6\Rightarrow B<5orA⩾4)orG=0$$$$\:\left({G}=\mathrm{0}\Rightarrow{I}=\mathrm{3}\Rightarrow{D}=\mathrm{6}\Rightarrow{E}=\mathrm{18}\right){X}…