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Algebra Difficulty 4.5 AIME Prove it Russia

Let a≠1a \neq 1 be a positive real number. Prove that the sequence x1,x2,…x_1, x_2, \dots, where xn=2n(2na−1)x_n = 2^n (2^n\sqrt{a} - 1), is strictly decreasing.

Solution

Пусть t=at = \sqrt{a}. Заметим, что t≠1t \neq 1. Тогда xn+1=2n+1(t−1)x_{n+1} = 2^{n+1}(t-1) и xn=2n(t2−1)x_n = 2^n(t^2 - 1). Следовательно,
xn−xn+1=2n(t2−1)−2n+1(t−1)==2n(t2−2t+1)=2n(t−1)2>0, \begin{align*} x_n - x_{n+1} &= 2^n(t^2 - 1) - 2^{n+1}(t-1) = \\ &= 2^n(t^2 - 2t + 1) = 2^n(t-1)^2 > 0, \end{align*}
что и требовалось доказать.

Замечание. При других подходах к решению неравенства могут выглядеть по-разному при a>1a > 1 (когда члены последовательности положительны) и при 0<a<10 < a < 1 (когда они отрицательны).

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