Problem 27 - Precalculus: Mathematical for Calculus

Question

F_{n}\( denotes the \)n\( th term of the Fibonacci sequence discussed in Section 11.1. Use mathematical induction to prove the statement. \)F_{3 n}\( is even for all natural numbers \)n$

Step-by-Step Solution

Answer

\(F_{3n}\) is even by induction for all natural numbers \(n\).

Step 1: Base Case

To prove the statement by induction, we start with the base case, checking for the smallest natural number, which is usually \(n=1\). In the Fibonacci sequence, \(F_3 = 2\). Since 2 is even, \(F_3\) is indeed even. Thus, the base case holds.

Step 2: Inductive Hypothesis

Assume that \(F_{3k}\) is even for some arbitrary natural number \(k\). This is our inductive hypothesis, and we aim to prove that \(F_{3(k+1)}\) is also even.

Step 3: Inductive Step

To prove the induction step, consider \(F_{3(k+1)} = F_{3k + 3}\). Using the Fibonacci sequence property, we know:\[F_{3k + 3} = F_{3k + 2} + F_{3k + 1}\]Additionally, since \(F_{3k + 2} = F_{3k + 1} + F_{3k}\) and \(F_{3k + 1} = F_{3k} + F_{3k - 1}\), substitute these to unfold \[F_{3k + 3} = (F_{3k+1} + F_{3k}) + F_{3k+1} = 2F_{3k+1} + F_{3k}\]We know \(F_{3k}\) is even by the inductive hypothesis, making the term \(F_{3k}\) even, and since the sum \(2F_{3k+1} + F_{3k}\) yields an even result for even \(F_{3k}\), \(F_{3(k+1)}\) is even.

Step 4: Conclusion

Since we've shown that the base case holds and the inductive step proves that if \(F_{3k}\) is even, then \(F_{3(k+1)}\) must also be even, we conclude by induction that \(F_{3n}\) is even for all natural numbers \(n\).

Key Concepts

Fibonacci Sequence

The Fibonacci sequence is a well-known sequence in mathematics, where each number is the sum of the two preceding ones, often starting with 0 and 1. This sequence has many interesting properties and appears in various natural phenomena. For instance, the sequence begins as follows:

• 0, 1, 1, 2, 3, 5, 8, 13, 21, 34...

Mathematically, the sequence can be defined by the recursive formula:\[ F_n = F_{n-1} + F_{n-2} \]with initial conditions \( F_0 = 0 \) and \( F_1 = 1 \). In this exercise, the focus is on the sequence at every third term, \( F_{3n} \), and proving that these terms are even numbers. The challenge is to use mathematical induction to demonstrate this intriguing property.

Base Case

In mathematical induction, the base case forms the foundation of the proof. It's crucial because it verifies that the statement holds true for the smallest integer, usually \( n=1 \) in most sequences.
In this exercise, we start with \( n = 1 \) and calculate the third term of the Fibonacci sequence, which is \( F_3 \). Since \( F_3 = 2 \) and 2 is an even number, the base case is verified. This step assures us that the formula we propose holds true at the beginning of the sequence.

Inductive Hypothesis

The inductive hypothesis is the second stage in the proof by induction. It's an assumption that the statement is true for a certain arbitrary natural number, say \( k \). This assumption is key to demonstrating that if something holds for \( k \), it must also hold for \( k+1 \).
In our problem, the inductive hypothesis is that \( F_{3k} \), the Fibonacci sequence's third term for any \( k \), is even. This assumption allows us to logically argue why the same property must apply to the subsequent term, \( F_{3(k+1)} \), using the properties of the sequence.

Inductive Step

The inductive step is where we prove that the truth of our statement for \( n = k \) implies its truth for \( n = k + 1 \). This is crucial in extending the induction argument across the entire range of natural numbers.
In this case, we approach this by examining \( F_{3(k+1)} \), which equals \( F_{3k + 3} \). Using the recursive nature of the Fibonacci sequence, we express \( F_{3k+3} \) in terms of previous terms:

• \( F_{3k + 3} = F_{3k + 2} + F_{3k + 1} \)

By substitution and simplification, we find:

• \( F_{3k + 3} = 2F_{3k+1} + F_{3k} \)

Since \( F_{3k} \) is even by our inductive hypothesis, and multiplication and addition of even numbers produce even results, \( F_{3k+3} \) is even. Thus, we've shown the property holds for \( k+1 \), concluding the proof.