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AlgebraQuestion and Answers: Page 63

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Let ξ be a positive Root of x^2 −2023x−1 Define a sequence ϕ_i such That ϕ_0 =1 ϕ_(n+1) =⌊ϕ_n ξ⌋, find The Remainder When ϕ_(2023 ) is divided by (√ϕ_2 )

$${Let}\:\xi\:{be}\:{a}\:{positive}\:{Root}\:{of}\:{x}^{\mathrm{2}} −\mathrm{2023}{x}−\mathrm{1} \\ $$$${Define}\:{a}\:{sequence}\:\varphi_{{i}} \:{such}\:{That}\:\varphi_{\mathrm{0}} =\mathrm{1} \\ $$$$\varphi_{{n}+\mathrm{1}} =\lfloor\varphi_{{n}} \xi\rfloor,\:{find}\:{The}\:{Remainder}\:{When}\:\varphi_{\mathrm{2023}\:} {is}\:{divided}\:{by}\:\sqrt{\varphi_{\mathrm{2}} } \\ $$

Question Number 196870 Answers: 1 Comments: 0

let b_i ∧ a_i >0 where i∈{1,2,3,...,n}& Σ_(i=1) ^n (b_i )=λ Prove that ((λ−(b_1 +b_2 ))/((b_1 +b_2 )))(a_1 +a_2 )+((λ−(b_1 +b_3 ))/((b_1 +b_3 )))(a_1 +a_3 )+....+((λ−(b_2 +b_3 ))/((b_2 +b_3 )))(a_2 +a_3 )+...((λ−(b_(n−1) +b_n ))/((b_(n−1) +b_n )))(a_(n−1) +a_n ) ≥(√(((n(n−1)(n−2)^2 )/4)×ΣΣ_(1≤i<j≤n) (a_i a_j )))

$${let}\:{b}_{{i}} \wedge\:{a}_{{i}} >\mathrm{0}\:{where}\:{i}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3},...,{n}\right\}\&\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({b}_{{i}} \right)=\lambda\:{Prove}\:{that} \\ $$$$\frac{\lambda−\left({b}_{\mathrm{1}} +{b}_{\mathrm{2}} \right)}{\left({b}_{\mathrm{1}} +{b}_{\mathrm{2}} \right)}\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} \right)+\frac{\lambda−\left({b}_{\mathrm{1}} +{b}_{\mathrm{3}} \right)}{\left({b}_{\mathrm{1}} +{b}_{\mathrm{3}} \right)}\left({a}_{\mathrm{1}} +{a}_{\mathrm{3}} \right)+....+\frac{\lambda−\left({b}_{\mathrm{2}} +{b}_{\mathrm{3}} \right)}{\left({b}_{\mathrm{2}} +{b}_{\mathrm{3}} \right)}\left({a}_{\mathrm{2}} +{a}_{\mathrm{3}} \right)+...\frac{\lambda−\left({b}_{{n}−\mathrm{1}} +{b}_{{n}} \right)}{\left({b}_{{n}−\mathrm{1}} +{b}_{{n}} \right)}\left({a}_{{n}−\mathrm{1}} +{a}_{{n}} \right) \\ $$$$\geqslant\sqrt{\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}×\underset{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} {\Sigma\Sigma}\left({a}_{{i}} {a}_{{j}} \right)} \\ $$$$ \\ $$

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