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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}}…

Question Number 195352 by sonukgindia last updated on 31/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 195342 by Matica last updated on 31/Jul/23 $$\:\:\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that}\:\:\forall{n}\:\in\:\mathbb{N}^{\ast} \:,\:\mathrm{4}^{{n}} \left({n}!\right)^{\mathrm{3}} \:<\:\left({n}+\mathrm{1}\right)^{\mathrm{3}{n}} \:. \\ $$$$\mathrm{2}.\:\mathrm{Solve}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{in}\:\mathbb{Z}^{\mathrm{2}} \:: \\ $$$$\:\:\:\:\:{a}./\:\:\mathrm{2}{x}^{\mathrm{3}} +{xy}−\mathrm{7}=\mathrm{0}\:, \\ $$$$\:\:\:\:\:{b}./\:\:{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{7}\right)\left({x}+\mathrm{8}\right)={y}^{\mathrm{2}} . \\ $$…

Question Number 195364 by Rodier97 last updated on 01/Aug/23 $$ \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{for}\:\mathrm{any}\:\mathrm{natural}\:\mathrm{number}\:{n},\: \\ $$$$\mathrm{the}\:\mathrm{natural}\:\mathrm{number}\:\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)^{{n}} +\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{{n}} \:\mathrm{is}\:\mathrm{divisible} \\ $$$$\mathrm{by}\:\mathrm{2}^{{n}} . \\ $$ Answered by Frix last…