Question Number 194147 by DAVONG last updated on 28/Jun/23 $$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)=? \\ $$ Answered by Frix last updated on 28/Jun/23 $$\frac{\pi^{\mathrm{2}}…
Question Number 194257 by tri26112004 last updated on 01/Jul/23 $${Know}\:{x},{y},{z}\:\in\:{R}^{+} \:{such}\:{that}: \\ $$$$\mathrm{2}{x}\:+\:\mathrm{4}{y}\:+\:\mathrm{7}{z}\:=\:\mathrm{2}{xyz} \\ $$$${Find}\:{Min}\left({x}+{y}+{z}\right)¿ \\ $$ Commented by Frix last updated on 01/Jul/23 $${x}=\mathrm{3}\wedge{y}=\frac{\mathrm{5}}{\mathrm{2}}\wedge{z}=\mathrm{2}\:\Rightarrow\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{15}}{\mathrm{2}}…
Question Number 194128 by tri26112004 last updated on 28/Jun/23 Answered by MM42 last updated on 28/Jun/23 $$\left.\:\:{a}\right)\:\:\:\:\underset{{n}=\mathrm{2023}} {\overset{\infty} {\sum}}\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} −\underset{{n}=\mathrm{0}} {\overset{\mathrm{2022}} {\sum}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}}…
Question Number 194255 by tri26112004 last updated on 01/Jul/23 $${Find}\:{value}\:{m}\:{so}\:{that}\:{the}\:{function}\: \\ $$$${y}=\mid{x}^{\mathrm{2}} −\mathrm{2}{mx}\mid−\mathrm{6}{x}\:{covariaties} \\ $$$$\:{on}\:{the}\:{interval}\:\left(\mathrm{1};\mathrm{4}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 194248 by Biteshwar last updated on 02/Jul/23 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Gyanashram}\:{classes} \\ $$$${weekly}\:{test}\:\:\:\:\:\:\:\:\:\:{by}−{Bittu}\:{sir}\:\: \\ $$$$\:\:\:\mathbb{CHEMISTRY}\:\:\mathbb{TEST} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Electrochemistry} \\ $$$$\mathrm{1}.\: \: \: \: \: \: \: \:…
Question Number 194241 by tri26112004 last updated on 01/Jul/23 Answered by mr W last updated on 02/Jul/23 $${assumed}\:{a},{b}\in{N} \\ $$$$\frac{\mathrm{1}}{{n}\left({n}+{a}\right)\left({n}+{b}\right)}=\frac{{A}}{{n}}+\frac{{B}}{{n}+{a}}+\frac{{C}}{{n}+{b}} \\ $$$$\left({A}+{B}+{C}\right){n}^{\mathrm{2}} +\left[\left(\mathrm{2}{a}+{b}\right){A}+{bB}\right]{n}+{abA}=\mathrm{1} \\ $$$${A}+{B}+{C}=\mathrm{0}…
Question Number 194237 by MrGHK last updated on 01/Jul/23 $$\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{evaluate}}\: \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{k}^{{n}} {n}!\left({zn}+\mathrm{1}\right)} \\ $$ Commented by TheHoneyCat last updated on 14/Jul/23…
Question Number 194236 by MrGHK last updated on 01/Jul/23 Answered by witcher3 last updated on 03/Jul/23 $$\mathrm{S}=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}!\left(\mathrm{zn}+\mathrm{1}\right)\mathrm{k}^{\mathrm{n}} } \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\int\left(\frac{−\mathrm{1}}{\mathrm{k}}\right)^{\mathrm{n}} .\frac{\mathrm{1}}{\mathrm{n}!}\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 194190 by tri26112004 last updated on 29/Jun/23 $${Explanation}\:{Why}: \\ $$$${While}\:{f}\left({ax}+{b}\right)+{f}\left({cx}+{d}\right)={ex}+{g} \\ $$$${then}\:{f}\left({x}\right)={Ax}^{\mathrm{2}} +{Bx}+{C}\:¿ \\ $$ Commented by Tinku Tara last updated on 30/Jun/23…
Question Number 194183 by tri26112004 last updated on 29/Jun/23 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{15}\right)\left({n}+\mathrm{30}\right)} \\ $$ Answered by ARUNG_Brandon_MBU last updated on 29/Jun/23 $$\frac{\mathrm{1}}{{n}\left({n}+\mathrm{15}\right)\left({n}+\mathrm{30}\right)}=\frac{\mathrm{1}}{\mathrm{450}{n}}−\frac{\mathrm{1}}{\mathrm{225}\left({n}+\mathrm{15}\right)}+\frac{\mathrm{1}}{\mathrm{450}\left({n}+\mathrm{30}\right)} \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{450}}\underset{{n}=\mathrm{0}} {\overset{\infty}…