Question Number 195419 by CrispyXYZ last updated on 02/Aug/23 $${x}\neq\mathrm{0}.\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{ix}^{{i}−\mathrm{1}} =? \\ $$ Answered by MathedUp last updated on 02/Aug/23 $${x}\neq\mathrm{0}.\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{ix}^{{i}−\mathrm{1}}…
Question Number 195413 by Calculusboy last updated on 02/Aug/23 Commented by Rasheed.Sindhi last updated on 02/Aug/23 $${x}^{{y}} +{y}^{{x}} =\mathrm{8}\Rightarrow{x}={y}=\mathrm{2} \\ $$$${x}+{y}={k}^{\mathrm{2}} \Rightarrow{k}^{\mathrm{2}} =\mathrm{2}+\mathrm{2}=\mathrm{4}\Rightarrow{k}=\pm\mathrm{2} \\ $$$${k}+\mathrm{1}=\pm\mathrm{2}+\mathrm{1}=\mathrm{3},−\mathrm{1}…
Question Number 195412 by Calculusboy last updated on 02/Aug/23 Answered by AST last updated on 02/Aug/23 $${v}_{\mathrm{2}} \left({a}^{\mathrm{2}^{{n}} } −\mathrm{1}\right)={v}_{\mathrm{2}} \left({a}−\mathrm{1}\right)+{n}+{v}_{\mathrm{2}} \left({a}+\mathrm{1}\right)−\mathrm{1}\geqslant{n}+\mathrm{2} \\ $$$$\Rightarrow\mathrm{2}^{{n}+\mathrm{2}} \mid{a}^{\mathrm{2}^{{n}}…
Question Number 195415 by Calculusboy last updated on 02/Aug/23 Answered by Frix last updated on 02/Aug/23 $$\mathrm{No}\:\mathrm{solution}\:\mathrm{exists}\:\mathrm{for}\:{x}\in\mathbb{C} \\ $$ Commented by Frix last updated on…
Question Number 195446 by sonukgindia last updated on 02/Aug/23 Answered by Frix last updated on 02/Aug/23 $$\sqrt{{a}+{b}\mathrm{i}}=\frac{\sqrt{{a}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}}{\:\sqrt{\mathrm{2}}}+\mathrm{sign}\:{b}\:\frac{\sqrt{−{a}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}}{\:\sqrt{\mathrm{2}}}\mathrm{i} \\ $$$$\sqrt{−\mathrm{110}+\mathrm{66i}}=\sqrt{\mathrm{11}\left(−\mathrm{5}+\sqrt{\mathrm{34}}\right)}+\mathrm{i}\sqrt{\mathrm{11}\left(\mathrm{5}+\sqrt{\mathrm{34}}\right)} \\ $$…
Question Number 195414 by tri26112004 last updated on 02/Aug/23 $${x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} }=¿ \\ $$ Answered by Frix last updated on 02/Aug/23 $$=\mathrm{2cosh}\:\left({n}\mathrm{ln}\:{x}\right) \\ $$ Commented…
Question Number 195443 by MM42 last updated on 02/Aug/23 $$\mathrm{10}^{\mathrm{10}} +\mathrm{10}^{\mathrm{10}^{\mathrm{2}} } +\mathrm{10}^{\mathrm{10}^{\mathrm{3}} } +…+\mathrm{10}^{\mathrm{10}^{\mathrm{10}} } \:\:\overset{\mathrm{7}} {\equiv}\:\:? \\ $$ Answered by BaliramKumar last updated…
Question Number 195411 by sonukgindia last updated on 02/Aug/23 Answered by MM42 last updated on 02/Aug/23 $${lim}_{{x}\rightarrow\infty} \:\frac{\mathrm{2}×\mathrm{2}^{{x}} −\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{9}^{{x}} +\mathrm{4}^{{x}} }{\mathrm{8}×\mathrm{4}^{{x}} +\mathrm{3}^{{x}} −\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{4}^{{x}} } \\…
Question Number 195436 by MrGHK last updated on 02/Aug/23 $$\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\boldsymbol{\mathrm{a}}\right)^{−\boldsymbol{\mathrm{n}}} }{\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right)^{\mathrm{2}} }\left(\boldsymbol{\psi}\left(\frac{\boldsymbol{\mathrm{n}}+\mathrm{4}}{\mathrm{2}}\right)−\boldsymbol{\psi}\left(\frac{{n}+\mathrm{3}}{\mathrm{2}}\right)\right)=??? \\ $$ Answered by witcher3 last updated on 02/Aug/23 $$\Psi\left(\mathrm{1}+\mathrm{x}\right)=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 195433 by sonukgindia last updated on 02/Aug/23 Answered by gatocomcirrose last updated on 02/Aug/23 $$\mathrm{2}\begin{vmatrix}{\mathrm{2}}&{−\mathrm{2}}&{\mathrm{1}}\\{\mathrm{x}}&{\mathrm{2}}&{−\mathrm{1}}\\{\mathrm{1}}&{−\mathrm{1}}&{−\mathrm{2}}\end{vmatrix}−\mathrm{3}\begin{vmatrix}{\mathrm{x}}&{−\mathrm{2}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{2}}&{−\mathrm{1}}\\{\mathrm{x}}&{−\mathrm{1}}&{−\mathrm{2}}\end{vmatrix}+ \\ $$$$+\mathrm{i}\begin{vmatrix}{\mathrm{x}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{x}}&{−\mathrm{1}}\\{\mathrm{x}}&{\mathrm{1}}&{−\mathrm{2}}\end{vmatrix}+\begin{vmatrix}{\mathrm{x}}&{\mathrm{2}}&{−\mathrm{2}}\\{\mathrm{2}}&{\mathrm{x}}&{\mathrm{2}}\\{\mathrm{x}}&{\mathrm{1}}&{−\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\left[−\mathrm{10}−\mathrm{5x}\right]−\mathrm{3}\left[−\mathrm{5x}−\mathrm{10}\right]+\mathrm{i}\left[−\mathrm{3x}^{\mathrm{2}} −\mathrm{x}+\mathrm{10}\right]+\left[\mathrm{x}^{\mathrm{2}} +\mathrm{2x}\right]=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}}…