Question Number 129869 by Bird last updated on 20/Jan/21 $${calculate}\:{lim}_{{n}\rightarrow+\infty} \sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{\left.\:\sqrt{\left({k}+{n}\right)\left({k}+\mathrm{1}+{n}\right.}\right)} \\ $$ Answered by mindispower last updated on 20/Jan/21 $$ \\ $$$$\frac{\mathrm{1}}{\left({k}+\mathrm{1}+{n}\right)}\leqslant\frac{\mathrm{1}}{\:\sqrt{{k}+{n}}.\sqrt{{k}+\mathrm{1}+{n}}}\leqslant\frac{\mathrm{1}}{{k}+{n}}…\mathrm{1}…
Question Number 64300 by mathmax by abdo last updated on 16/Jul/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 129403 by bemath last updated on 15/Jan/21 $$\:\mathrm{f}\left(\mathrm{n}\right)\:=\:\mathrm{4f}\left(\mathrm{n}−\mathrm{1}\right)\:−\mathrm{4f}\left(\mathrm{n}−\mathrm{2}\right)\:+\:\mathrm{n}^{\mathrm{2}} \: \\ $$$$\mathrm{and}\:\mathrm{f}\left(\mathrm{0}\right)=\mathrm{2}\:,\:\mathrm{f}\left(\mathrm{1}\right)=\mathrm{5}\: \\ $$ Commented by talminator2856791 last updated on 15/Jan/21 $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{question}? \\ $$…
Question Number 129383 by mathmax by abdo last updated on 15/Jan/21 $$\mathrm{find}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{u}_{\mathrm{n}} \:\mathrm{wish}\:\mathrm{verify}\:\mathrm{u}_{\mathrm{n}−\mathrm{p}} +\mathrm{u}_{\mathrm{n}+\mathrm{p}} =\frac{\left(−\mathrm{1}\right)^{\mathrm{p}} }{\mathrm{n}+\mathrm{p}} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{n}\geqslant\mathrm{p} \\ $$ Terms of Service Privacy Policy…
Question Number 129336 by bramlexs22 last updated on 15/Jan/21 $$\mathrm{If}\:{a},{b}\:\mathrm{and}\:{c}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{3}} −\mathrm{21}{x}−\mathrm{35}\:=\:\mathrm{0}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\:\left({a}−{b}\right)\left({c}−{a}\right)\left({b}−{c}\right)\:? \\ $$ Commented by MJS_new last updated on 22/Jan/21 $${x}^{\mathrm{3}}…
Question Number 129213 by bramlexs22 last updated on 13/Jan/21 $$\:\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{2x}+\mathrm{1}}\:\mathrm{and}\:\frac{\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)−\mathrm{1}}{\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)+\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{then}\:\mathrm{x}\:=? \\ $$ Answered by liberty last updated on 13/Jan/21 $$\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}}…
Question Number 63665 by mathmax by abdo last updated on 07/Jul/19 $${find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$ Commented by mathmax by abdo last…
Question Number 63651 by mathmax by abdo last updated on 06/Jul/19 $${let}\:{S}_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{e}^{−{k}} {sin}\left({k}^{\mathrm{2}} {x}\right) \\ $$$$\left.\mathrm{1}\right)\:{determine}\:\mathrm{2}\:{sequence}\:\:{U}_{{n}} \left({x}\right)\:{and}\:{V}_{{n}} \left({x}\right)\:{wich}\:{verify}\:{U}_{{n}} \leqslant\:{S}_{{n}} \leqslant\:{V}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{let}\:\:{S}\:={lim}_{{n}\rightarrow+\infty}…
Question Number 63560 by mathmax by abdo last updated on 05/Jul/19 $${developp}\:{at}\:{laurent}\:{series} \\ $$$$\left.\mathrm{1}\right)\:{f}\left({z}\right)\:=\frac{\mathrm{1}}{{z}−\mathrm{2}} \\ $$$$\left.\mathrm{2}\right){g}\left({z}\right)\:=\frac{\mathrm{3}}{{z}^{\mathrm{2}} −\mathrm{3}{z}\:+\mathrm{2}} \\ $$$$\left.\mathrm{3}\right){h}\left({z}\right)\:=\frac{\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{4}} \\ $$ Commented by mathmax…
Question Number 129052 by oustmuchiya@gmail.com last updated on 12/Jan/21 $${Let}\:\boldsymbol{\mathrm{p}}\:{and}\:\boldsymbol{\mathrm{Q}}\:{be}\:{points}\:{on}\:{the}\:{curve} \\ $$$$\boldsymbol{{y}}=\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{x}}\:{while}\:\boldsymbol{{x}}=\mathrm{2}\:{and}\:\boldsymbol{{x}}=\mathrm{2}+\boldsymbol{{h}} \\ $$$$\boldsymbol{{respectively}}.\:\boldsymbol{{E}}{press}\:{the}\:{gradient} \\ $$$${of}\:\boldsymbol{\mathrm{P}}{Q}\:{in}\:{terms}\:{of}\:\boldsymbol{{h}}. \\ $$ Commented by benjo_mathlover last updated on…