Question Number 144767 by mnjuly1970 last updated on 29/Jun/21 | ||
Answered by Rasheed.Sindhi last updated on 29/Jun/21 | ||
$$\left({x}^{\mathrm{3}} −\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\right)^{{m}} ={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +...+{a}_{{n}} {x}^{{n}} \\ $$$$\frac{{a}_{\mathrm{0}} +{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +...+{a}_{{n}} }{{n}+\mathrm{1}}=\mathrm{0}.\mathrm{2}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${Let}\:{x}=\mathrm{1} \\ $$$$\left(−\mathrm{1}\right)^{{m}} ={a}_{\mathrm{0}} +{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +...+{a}_{{n}} =\frac{{n}+\mathrm{1}}{\mathrm{5}} \\ $$$${m}\in\mathbb{O}\rightarrow\frac{{n}+\mathrm{1}}{\mathrm{5}}=−\mathrm{1}\Rightarrow{n}=−\mathrm{6}\left({rejected}\right) \\ $$$${m}\notin\mathbb{O} \\ $$$${m}\in\mathbb{E}\rightarrow\:\frac{{n}+\mathrm{1}}{\mathrm{5}}=\mathrm{1}\Rightarrow{n}=\mathrm{4} \\ $$$${a}_{\mathrm{0}} +{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +{a}_{\mathrm{4}} =\frac{\mathrm{4}+\mathrm{1}}{\mathrm{5}}=\mathrm{1} \\ $$$$....... \\ $$$$..... \\ $$ | ||
Answered by mr W last updated on 29/Jun/21 | ||
$$\left({x}^{\mathrm{3}} −\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\right)^{{m}} ={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +...+{a}_{{n}} {x}^{{n}} \\ $$$${x}=\mathrm{1}: \\ $$$$\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}}\right)^{{m}} ={a}_{\mathrm{0}} +{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +...+{a}_{{n}} =\frac{{n}+\mathrm{1}}{\mathrm{5}} \\ $$$$\left(−\mathrm{1}\right)^{{m}} =\frac{{n}+\mathrm{1}}{\mathrm{5}} \\ $$$$\Rightarrow{n}=\mathrm{4},{m}=\mathrm{4} \\ $$$$\left({x}^{\mathrm{3}} −\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\right)^{\mathrm{4}} =\frac{\mathrm{16}}{{x}^{\mathrm{8}} }−\frac{\mathrm{32}}{{x}^{\mathrm{3}} }+\mathrm{24}{x}^{\mathrm{2}} −\mathrm{8}{x}^{\mathrm{7}} +{x}^{\mathrm{12}} \\ $$$${a}_{\mathrm{2}} =\mathrm{24} \\ $$ | ||
Commented by Rasheed.Sindhi last updated on 30/Jun/21 | ||
$$\mathcal{N}\:\overset{\bullet} {\shortmid}\:\subset\:\in\:\boldsymbol{\mathrm{Sir}}! \\ $$ | ||
Commented by mnjuly1970 last updated on 30/Jun/21 | ||
$$\:\:{thx}\:{mr}\:{W}... \\ $$ | ||