Problem
ALG-B2-M01-P028 A decreasing root sequence
Let \(a>0\), \(a\ne1\), and \[x_n=2^n\left(\sqrt[2^n]{a}-1\right).\] Prove that the sequence \(x_1,x_2,\ldots\) is strictly decreasing.
Hint 1. Relate \(x_n\) and \(x_{n+1}\) using one variable.
Hint 2. Let \(t=\sqrt[2^{n+1}]{a}\), so \(\sqrt[2^n]{a}=t^2\).
Put \(t=\sqrt[2^{n+1}]{a}\). Then \(t\ne1\), \(\sqrt[2^n]{a}=t^2\), and \[x_n=2^n(t^2-1),\qquad x_{n+1}=2^{n+1}(t-1).\] Hence \[x_n-x_{n+1}=2^n(t^2-1)-2^{n+1}(t-1)=2^n(t-1)^2>0.\] Thus \(x_n>x_{n+1}\) for every \(n\), so the sequence is strictly decreasing.
A. Source analysis. Main objects: inequalities, order, an extremal element, or an invariant. The obvious first move usually gives only a local estimate. The hidden observation is to choose the right nondecreasing quantity, or to add/multiply inequalities only after signs are controlled. The needed step is an ordering, an invariant, a product transformation, or a boundary case.
F. Difficulty justification. Regional level 7: the hidden step is expressing neighbouring terms through the same root.
G. Why this is not a one-step exercise. The roots cannot be compared directly without the right substitution.