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Algebra Difficulty 4.7 AIME Find the answer

6. For any integer n(n⩾2)n(n \geqslant 2), satisfying an=a+1,b2n=b+3aa^{n}=a+1, b^{2 n}=b+3 a
the size relationship of the positive numbers aa and bb is:

Pick one

Solution

6.A.

From an=a+1>1a^{n}=a+1>1, we know a>1a>1.
Also, b2n=b+3a>3a>3b^{2 n}=b+3 a>3 a>3, thus, b>1b>1.
Therefore, aa and bb are both greater than 1.
Since a2n−b2na^{2 n}-b^{2 n}
=(a+1)2−(b+3a)=a2−a−b+1, and a2n−b2n=(a−b)(a2n−1+a2n−2b+⋯+b2n−1), \begin{array}{l} =(a+1)^{2}-(b+3 a)=a^{2}-a-b+1, \\ \text { and } a^{2 n}-b^{2 n} \\ =(a-b)\left(a^{2 n-1}+a^{2 n-2} b+\cdots+b^{2 n-1}\right), \end{array}

then a2−a−b+1a^{2}-a-b+1
=(a−b)(a2n−1+a2n−2b+⋯+b2n−1).  =(a-b)\left(a^{2 n-1}+a^{2 n-2} b+\cdots+b^{2 n-1}\right) \text {. }

Therefore, a2−a−b+1a−b>1\frac{a^{2}-a-b+1}{a-b}>1.
Hence a2−a−b+1−a+ba−b>0\frac{a^{2}-a-b+1-a+b}{a-b}>0, which means
(a−1)2a−b>0.  \frac{(a-1)^{2}}{a-b}>0 \text {. }

Thus, a>ba>b.
In summary, a>b>1a>b>1.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.