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Challenge Problem Paper Creases If a sheet of paper is folded in half by folding the top edge down to the bottom edge, one crease will result. If the folded paper is folded in the same manner, the result is three creases. With each fold, the number of creases can be defined recursively by \(c_{1}=1, c_{n+1}=2 c_{n}+1\) (a) Find the number of creases for \(n=3\) and \(n=4\) folds. (b) Use the given information and your results from part (a) to find a formula for the number of creases after \(n\) folds, \(c_{n}\), in terms of the number of folds alone. (c) Use the Principle of Mathematical Induction to prove that the formula found in part (b) is correct for all natural numbers. (d) Tosa Tengujo is reportedly the world's thinnest paper with a thickness of \(0.02 \mathrm{~mm}\). If a piece of this paper could be folded 25 times, how tall would the stack be?

Step by step solution

01

Understanding the Recursive Relation

Given: The number of creases after each fold is recursively defined as \[c_{1} = 1, \; c_{n+1} = 2c_{n} + 1\] First, we will use this recursive formula to find the number of creases for \(n=3\) and \(n=4\).

02

Calculate Number of Creases for n=3 and n=4

For \(n=3\), we calculate as follows: \[c_{2} = 2c_{1} + 1 = 2(1) + 1 = 3\]\[c_{3} = 2c_{2} + 1 = 2(3) + 1 = 7\]Similarly, for \(n=4\):\[c_{4} = 2c_{3} + 1 = 2(7) + 1 = 15\] Therefore, the number of creases for \(n=3\) folds is 7 and for \(n=4\) folds is 15.

03

Formulate the General Expression

Using the pattern observed from the recursive calculation: \(c_1 = 1\)\(c_2 = 3 = 2^2 - 1\)\(c_3 = 7 = 2^3 - 1\)\(c_4 = 15 = 2^4 - 1\)We generalize this and define the formula for an arbitrary number of folds \(n\). The pattern suggests:\[c_{n} = 2^n - 1\]

04

Prove the Formula using Mathematical Induction

Base Case: For \(n=1\), we see \[c_1 = 1 = 2^1 - 1\] Induction Step: Assume the formula \(c_k = 2^k - 1\) is true for some \(k\). We need to show that it holds for \(k+1\). Using the recursive formula: \[c_{k+1} = 2c_k + 1\] Substitute the induction hypothesis: \[c_{k+1} = 2(2^k - 1) + 1 = 2^{k+1} - 2 + 1 = 2^{k+1} - 1\] This completes the induction proof. Therefore, the formula \(c_n = 2^n - 1\) holds for all natural numbers \(n\).

05

Calculate the Height of the Stack for 25 Folds

Given the thickness of the paper is \(0.02 \, \text{mm}\), the height of the paper stacked after 25 folds is calculated based on folding. Each fold doubles the number of layers: \[h_{25} = 0.02 \times 2^{25} \, \text{mm}\] Using a calculator or estimating: \[2^{25} = 33,554,432\] Therefore, the height is: \[h_{25} = 0.02 \times 33,554,432 = 671,088.64 \, \text{mm} \] Converting to meters: \[671,088.64 \, \text{mm} = 671.09 \, \text{meters}\]

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Key Concepts

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

Recurrence Relations

A recurrence relation is a formula that defines each term of a sequence from the preceding terms. This concept is crucial for understanding recursive sequences. In our problem, the number of creases after each fold follows a recurrence relation: \[ c_{n+1} = 2c_n + 1 \] Here:

• \( c_{1} \) is the base case, with 1 crease.
• \( c_{n+1} \) depends on the previous number of creases \( c_n \).

By applying this relation iteratively, we can find the number of creases for any number of folds \( n \). For example, \( c_{2} = 3 \) and \( c_{3} = 7 \), showing how each term builds off the previous term. This type of relationship helps simplify complex calculations by breaking them down into manageable steps.

Mathematical Induction

Mathematical induction is a powerful proof technique used to establish the truth of an infinite number of cases. To use induction for our formula, we need two key steps:

• **Base Case**: Verify the formula for \( n = 1 \), which we did as \( c_{1} = 1 \).
• **Inductive Step**: Assume the formula is true for some \( n = k \), and then prove it for \( n = k+1 \).

In our problem, we assume \( c_k = 2^k - 1 \) is true. From the recurrence relation, we calculate: \[ c_{k+1} = 2c_k + 1 = 2(2^k - 1) + 1 = 2^{k+1} - 1 \] By confirming both steps, we prove the formula \( c_n = 2^n - 1 \) is correct for all natural numbers \( n \). Induction helps validate recursive definitions rigorously, ensuring they hold universally.

Exponential Growth

Exponential growth describes quantities that increase rapidly over time. In this paper-folding problem, both the number of creases and the paper's thickness after each fold grow exponentially. For creases, the formula \( c_n = 2^n - 1 \) shows exponential growth with each fold doubling the number of previous creases plus one, illustrating rapid increase. For the thickness, each fold doubles the paper layers. Starting with 0.02 mm, after 25 folds: \[ h_{25} = 0.02 \times 2^{25} = 671,088.64 \text{ mm} = 671.09 \text{ meters} \] This exponential growth highlights how quickly quantities can surge, turning a thin paper into an astonishingly tall stack in just a few folds. Recognizing exponential behavior is key in various fields, from population dynamics to finance.

Folding Geometry

Folding geometry explores the mathematical attributes and principles involved in folding objects, such as paper. Each fold influences the number of creases exponentially due to the recursive nature of the folding. Initially, one fold creates one crease. But if you fold the paper again, subsequent folds double the creases formed plus one more. This creates a predictable pattern, since:

• First fold: 1 crease.
• Second fold: 3 creases.
• Third fold: 7 creases.

The geometrical patterns generated by folding follow strict mathematical rules, as seen in our recurrence relation. Additionally, the thickness of the paper stack grows exponentially, obeying the same doubling rule. Studying folding geometry reveals surprising and complex outcomes from simple iterative processes, and can be applied to various scientific and engineering disciplines to solve real-world problems.