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2

Part of 1992 Romania Team Selection Test

Problems(2)

x_{n+2 }= (a^4 +4a^2 +2)x_{n+1} -x_n -2a^2, perfext square, matrix

Source: Romania BMO TST 1992 p2

2/19/2020
For a positive integer aa, define the sequence (xnx_n) by x1=x2=1x_1 = x_2 = 1 and xn+2=(a4+4a2+2)xn+1−xn−2a2x_{n+2 }= (a^4 +4a^2 +2)x_{n+1} -x_n -2a^2 , for n ≥1\ge 1. Show that xnx_n is a perfect square and that for n>2n > 2 its square root equals the first entry in the matrix (a2+1aa1)n−2\begin{pmatrix} a^2+1 & a \\ a & 1 \end{pmatrix}^{n-2}
linear algebramatrixPerfect SquareSequence
Romanian TST 1992

Source: Romanian TST 1992 - Day 2 - Problem 2

4/9/2012
Let a1,a2,...,ak a_1, a_2, ..., a_k be distinct positive integers such that the 2k2^k sums ∑i=1kϵiai\displaystyle\sum\limits_{i=1}^{k}{\epsilon_i a_i}, ϵi∈{0,1}\epsilon_i\in\left\{0,1\right\} are distinct. a) Show that 1a1+1a2+...+1ak≤2(1−2−k) \dfrac{1}{a_1}+\dfrac{1}{a_2}+...+\dfrac{1}{a_k}\le2(1-2^{-k}) ; b) Find the sequences (a1,a2,...,ak)(a_1,a_2,...,a_k) for which the equality holds.
Șerban Buzețeanu
inequalitiesinequalities proposed