Maths Olympiad Prep

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Number theory Difficulty 6.1 National olympiad Prove it

Example 9 Arrange a strip of width 1 using rectangles of width 1 and lengths a1,a2,a3,(a11)a_{1}, a_{2}, a_{3}, \cdots \left(a_{1} \neq 1\right), which are placed tightly and without overlapping. Starting from the second rectangle, each rectangle is similar but not congruent to the rectangle formed by the previously arranged rectangles. After placing the first nn rectangles, the length of the strip covered is denoted as SnS_{n}. Does there exist a real number that is not exceeded by SnS_{n}? Prove your conclusion.

Solution

Proof As shown in Figure 13313-3, according to the problem, we have 1:Sn=an+1:11: S_{n}=a_{n+1}: 1, so an+1=1Sna_{n+1}=\frac{1}{S_{n}}. Since Sn+1=Sn+an+1S_{n+1}=S_{n}+a_{n+1}, it follows that for all n(nN)n(n \in \mathbf{N}), we have Sn+1=Sn+1SnS_{n+1}=S_{n}+\frac{1}{S_{n}}.
Therefore, Sn+12=(Sn+1Sn)2>Sn2+2S_{n+1}^{2}=\left(S_{n}+\frac{1}{S_{n}}\right)^{2}>S_{n}^{2}+2
which means Sn+12Sn2>2n=1,2,S_{n+1}^{2}-S_{n}^{2}>2 \quad n=1,2, \cdots
Summing up, we get Sn+12S12>2nS_{n+1}^{2}-S_{1}^{2}>2 n, i.e., Sn+12>2n+S12>2nS_{n+1}^{2}>2 n+S_{1}^{2}>2 n

Thus, the sequence {Sn2}\left\{S_{n}^{2}\right\} is monotonically increasing and unbounded, and so is {Sn}\left\{S_{n}\right\}. Therefore, there does not exist a real number that is not exceeded by SnS_{n}.

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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.