91Ó°ÊÓ

128\. Let \(a_{n}\) be the number of different strings of length \(n\) that can be formed from the symbols \(\mathrm{X}\) and \(\mathrm{O}\) with the restriction that a string may not consist of identical smaller strings. For example, XXXX and XOXO are not allowed. The possible strings of length 4 are XXXO, XXOX, XXOO, XOXX, XOOX, XOOO, OXXX, OXXO, OXOO, OOXX, OOXO, OOOX, so \(a_{4}=12\). Here is a table showing \(a_{n}\) and the ratio \(\frac{a_{n+1}}{a_{n}}\) for \(n=1,2, \ldots, 13\). \begin{tabular}{cccccccccccccc} \(n\) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\ \hline\(a_{n}\) & 2 & 2 & 6 & 12 & 30 & 54 & 126 & 240 & 504 & 990 & 2046 & 4020 & 8190 \\ \(\frac{a_{n+1}}{a_{n}}\) & 1 & 3 & 2 & \(2.5\) & \(1.8\) & \(2.33\) & \(1.90\) & \(2.1\) & \(1.96\) & \(2.07\) & \(1.96\) & \(2.03\) & \(1.98\) \end{tabular}

Step by step solution

01

- Analyzing the given data

Based on the table, analyze the values of \(a_n\) and the ratios \(\frac{a_{n+1}}{a_n}\) for \(n = 1, 2, \ldots, 13\). Notice that the number of different strings \(a_n\) increases as \(n\) increases. This increase is not uniform, but rather it varies with each step.

02

- Observing pattern in ratio

Observe how the ratios \(\frac{a_{n+1}}{a_n}\) fluctuate but revolve around an approximate value of 2 as \(n\) increases. The ratios generally fluctuate slightly around this average value.

03

- Generalizing the behavior

Given the data, when \(n\) increases, \(a_n\) tends to grow at a rate close to doubling. This observation leads to the inference that the ratios converge to approximately 2 as \(n\) becomes large.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

String Patterns

In combinatorics, string patterns refer to sequences created using a set of symbols. Here, we're using the symbols X and O. We're interested in finding the number of valid strings of a certain length, which don't consist of repeating smaller substrings. For instance, XXXX and XOXO are not allowed because they're composed of repeating smaller units (XXXX is a repeat of X, and XOXO is a repeat of XO). To solve this, we systematically generate and count all valid strings while ensuring they meet the given restriction. This approach can be complex as the length of the string increases, which is why understanding these patterns is important for combinatorial problems.

Growth Rates

The concept of growth rates helps us understand how quickly quantities increase. In our problem, the number of valid strings, represented as \(a_n\), grows as the length of the strings, \(n\), increases. When analyzing the table of \(a_n\) values, it's clear that the growth is not constant. Instead, it fluctuates based on certain patterns or rules. To better understand how quickly \(a_n\) grows, we look at the ratios \(\frac{a_{n+1}}{a_n}\), which tell us the change in the number of valid strings from one length to the next. These ratios tend to hover around 2, indicating that the number of valid strings roughly doubles as the length increases.

Recurrence Relations

Understanding recurrence relations is key to solving problems in combinatorics. A recurrence relation is an equation that defines a sequence based on previous terms. For example, if we know \(a_1\) and \(a_2\), we might use a recurrence relation to find \(a_3\), \(a_4\), and so on. In our problem, although the recurrence relation isn't explicitly given, the pattern seen in \(\frac{a_{n+1}}{a_n}\) hints that \(a_n\) can be approximated by doubling the previous term or using a similar growth factor. Recognizing these relationships allows us to predict the behavior of the sequence for larger values of \(n\), thereby effectively solving combinatorial challenges more efficiently.