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AlgebraQuestion and Answers: Page 53

Question Number 197125 Answers: 1 Comments: 1

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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 )

$L e t ξ b e a p o s i t i v e R o o t o f x^{2} - 2023 x - 1$ $D e f i n e a s e q u e n c e φ_{i} s u c h T h a t φ_{0} = 1$ $φ_{n + 1} = ⌊ φ_{n} ξ ⌋, f i n d T h e R e m a i n d e r W h e n φ_{2023} i s d i v i d e d b y \sqrt{φ_{2}}$

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 )))

$l e t b_{i} \land a_{i} > 0 w h e r e i \in {1, 2, 3, . . ., n} & \underset{i = 1}{\sum^{n}} (b_{i}) = λ P r o v e t h a t$ $\frac{λ - (b_{1} + b_{2})}{(b_{1} + b_{2})} (a_{1} + a_{2}) + \frac{λ - (b_{1} + b_{3})}{(b_{1} + b_{3})} (a_{1} + a_{3}) + . . . . + \frac{λ - (b_{2} + b_{3})}{(b_{2} + b_{3})} (a_{2} + a_{3}) + . . . \frac{λ - (b_{n - 1} + b_{n})}{(b_{n - 1} + b_{n})} (a_{n - 1} + a_{n})$ $⩾ \sqrt{\frac{n (n - 1) {(n - 2)}^{2}}{4} \times \underset{1 ⩽ i < j ⩽ n}{Σ Σ} (a_{i} a_{j})}$