Question Number 87179 by peter frank last updated on 03/Apr/20 Commented by peter frank last updated on 03/Apr/20 $${qn}\:\mathrm{2} \\ $$ Commented by Rio Michael…
Question Number 152670 by liberty last updated on 31/Aug/21 Answered by Olaf_Thorendsen last updated on 31/Aug/21 $${x}_{{n}} \:=\:{x}_{{n}−\mathrm{1}} +{x}_{{n}−\mathrm{2}} \\ $$$${x}_{{n}} \:−{x}_{{n}−\mathrm{1}} −{x}_{{n}−\mathrm{2}} \:=\:\mathrm{0} \\…
Question Number 21502 by 786786AM last updated on 25/Sep/17 $$\mathrm{In}\:\mathrm{an}\:\mathrm{A}.\mathrm{P},\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{p}\:\mathrm{terms}\:\mathrm{is}\:\mathrm{q}\:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{q}\:\mathrm{terms}\:\mathrm{is}\:\mathrm{p}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\left(\mathrm{p}+\mathrm{q}\right)\:\mathrm{terms}\:\mathrm{is}\:−\left(\mathrm{p}+\mathrm{q}\right). \\ $$ Answered by $@ty@m last updated on 25/Sep/17 $${ATQ} \\ $$$$\boldsymbol{\mathrm{S}}_{{p}} ={q}…
Question Number 152546 by liberty last updated on 29/Aug/21 Answered by qaz last updated on 29/Aug/21 $$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}\centerdot\mathrm{3}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{7}\centerdot\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{11}\centerdot\mathrm{3}^{\mathrm{5}} }+\frac{\mathrm{1}}{\mathrm{13}\centerdot\mathrm{3}^{\mathrm{6}} }−\frac{\mathrm{1}}{\mathrm{17}\centerdot\mathrm{3}^{\mathrm{8}} }−\frac{\mathrm{1}}{\mathrm{19}\centerdot\mathrm{3}^{\mathrm{9}} }+… \\ $$$$=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{7}\centerdot\mathrm{3}^{\mathrm{3}}…
Question Number 152496 by mathdanisur last updated on 28/Aug/21 Answered by qaz last updated on 29/Aug/21 $$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{x}^{\mathrm{4n}} }{\left(\mathrm{4n}\right)!}=\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}!}+\frac{\mathrm{x}^{\mathrm{8}} }{\mathrm{8}!}+\frac{\mathrm{x}^{\mathrm{12}} }{\mathrm{12}!}+… \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}}…
Question Number 86855 by jagoll last updated on 01/Apr/20 $$\mathrm{If}\:\mathrm{a},\mathrm{b}\:,\mathrm{c}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{x}^{\mathrm{3}} +\mathrm{6x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{3}\:=\:\mathrm{0}\:.\:\mathrm{find}\:\mathrm{the}\: \\ $$$$\mathrm{equation}\:\mathrm{with}\:\mathrm{roots}\:\mathrm{a}+\mathrm{b}\:,\:\mathrm{b}+\mathrm{c}\:,\:\mathrm{a}+\mathrm{c}\:? \\ $$ Answered by john santu last updated on…
Question Number 86791 by TawaTawa1 last updated on 31/Mar/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\:\:\:\:\mathrm{x}\:\:\equiv\:\:\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\:\:\:\:\mathrm{x}\:\:\equiv\:\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$$$\:\:\:\:\mathrm{x}\:\:\equiv\:\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{3}\right) \\ $$ Answered by mr W last updated on…
Question Number 86786 by M±th+et£s last updated on 31/Mar/20 $${show}\:{that} \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left(\mathrm{1}−{n}^{\mathrm{2}} \right)^{\mathrm{2}} }=\mathrm{0}.\mathrm{2999} \\ $$ Commented by Prithwish Sen 1 last…
Question Number 21154 by youssoufab last updated on 14/Sep/17 $${prove}:\:\forall{n}\in\mathbb{N}^{\ast} ,\forall\left({a},{b}\right)\in\mathbb{C}^{\mathrm{2}} ,\:{a}^{\mathrm{2}{n}+\mathrm{1}} +{b}^{\mathrm{2}{n}+\mathrm{1}} = \\ $$$$\underset{{k}={o}} {\overset{\mathrm{2}{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} {a}^{{k}} {b}^{\mathrm{2}{n}−{k}} \\ $$ Commented by alex041103…
Question Number 86672 by M±th+et£s last updated on 30/Mar/20 $${show}\:{proofs}\:{by}\:{induction},{that} \\ $$$$\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +….+{x}_{{n}} }{{n}}\geqslant\left({x}_{\mathrm{1}} {x}_{\mathrm{2}} ….{x}_{{n}} \right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$\forall{n}=\mathrm{2}^{{k}} ,{k}>\mathrm{1}\:{and}\:\left({x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,{x}_{\mathrm{3}} ,…..{x}_{{n}} \right)>\mathrm{0}.…