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Category: Algebra

1-9-1-3-2-9-1-3-4-9-1-3-

Question Number 103286 by bemath last updated on 14/Jul/20 $$\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{9}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}}{\mathrm{9}}}+\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{4}}{\mathrm{9}}}\:=\:? \\ $$ Commented by bemath last updated on 14/Jul/20 $${yes}.\:{thank}\:{both} \\ $$ Answered by floor(10²Eta[1])…

k-0-1-k-k-4-k-2-1-

Question Number 103278 by bachamohamed last updated on 13/Jul/20 $$\:\:\:\:\:\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\boldsymbol{{k}}!\left(\boldsymbol{{k}}^{\mathrm{4}} +\boldsymbol{{k}}^{\mathrm{2}} +\mathrm{1}\right)}=? \\ $$ Commented by Dwaipayan Shikari last updated on 13/Jul/20 $$\mathrm{By}\:\mathrm{ratio}\:\mathrm{test}\:\mathrm{it}\:\mathrm{converges}\:\mathrm{sir}!…

Question-168813

Question Number 168813 by safojontoshtemirov last updated on 18/Apr/22 Answered by aleks041103 last updated on 18/Apr/22 $${We}\:{can}\:{add}\:{and}/{or}\:{subtract}\:{different} \\ $$$${rows}\:{from}\:{each}\:{other}\:{without}\:{changing} \\ $$$${the}\:{determinant}. \\ $$$$\Rightarrow{d}_{{n}} =\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\ldots}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{2}}&{\ldots}&{\mathrm{2}}&{\mathrm{2}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}&{\ldots}&{\mathrm{3}}&{\mathrm{3}}\\{\vdots}&{\vdots}&{\vdots}&{\ddots}&{\vdots}&{\vdots}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}&{\ldots}&{{n}−\mathrm{1}}&{{n}−\mathrm{1}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}&{\ldots}&{{n}−\mathrm{1}}&{{n}}\end{vmatrix} \\…

Question-103260

Question Number 103260 by 675480065 last updated on 13/Jul/20 Answered by Rio Michael last updated on 13/Jul/20 $$\:\begin{cases}{{u}_{\mathrm{0}} \:=\:\mathrm{1}}\\{{u}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{2}{u}_{{n}} {v}_{{n}} }{{u}_{{n}} +\:{v}_{{n}} }}\end{cases}\:\mathrm{and}\:\begin{cases}{{v}_{\mathrm{0}} \:=\:\mathrm{2}}\\{{v}_{{n}+\mathrm{1}\:}…

i-1-i-

Question Number 103253 by Study last updated on 13/Jul/20 $$\:\sqrt[{{i}}]{{i}}=????? \\ $$ Answered by Dwaipayan Shikari last updated on 13/Jul/20 $${i}^{\frac{\mathrm{1}}{{i}}} ={i}^{−{i}} \:\:\:\:\left(\frac{\mathrm{1}}{{i}}=\frac{{i}}{{i}^{\mathrm{2}} }=−{i}\right) \\…