Question

Prove by induction that $$F_{n}=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}$$ is a closed form for the Fibonacci sequence.

Short answer

The formula is valid for all Fibonacci numbers by induction.

Step-by-step solution

Understanding the Problem

We need to use mathematical induction to prove that the given formula \( F_n = \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{n+1} - \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n+1} \) represents the Fibonacci sequence for all natural numbers \( n \), where \( F_n \) is defined as \( F_0 = 0 \) and \( F_1 = 1 \), and \( F_{n} = F_{n-1} + F_{n-2} \) for \( n \geq 2 \).

Base Case

For the base cases, compute \( F_0 \) and \( F_1 \) using the formula.\- Compute \( F_0 = \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{1} - \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{1} \).\- Similarly, compute \( F_1 = \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{2} - \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{2} \).\Show that they simplify to the Fibonacci sequence values: \( F_0 = 0 \) and \( F_1 = 1 \).

Inductive Hypothesis

Assume the formula holds for some integer \( k \), i.e., \( F_k = \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{k+1} - \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{k+1} \) and \( F_{k-1} = \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{k} - \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{k} \).

Inductive Step

Using the inductive hypothesis, compute \( F_{k+1} = F_k + F_{k-1} \). Substitute the expressions for \( F_k \) and \( F_{k-1} \) in terms of the closed-form formula. Simplify the expression to show:\[\begin{align*}F_{k+1} &= \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{k+2} - \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{k+2}.\end{align*}\]This matches the form of \( F_{k+1} \), thereby completing the inductive step.

Conclusion

Since the base case holds and the inductive step has been proven, by the principle of mathematical induction, the closed formula \( F_n = \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{n+1} - \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n+1} \) is valid for all natural numbers \( n \) in representing the Fibonacci sequence.

Key concepts

Fibonacci Sequence

The Fibonacci sequence is one of the most famous sequences in mathematics. It is defined by the relation:

• \( F_0 = 0 \)
• \( F_1 = 1 \)
• For \( n \geq 2 \), \( F_n = F_{n-1} + F_{n-2} \)

This sequence was introduced to the West in 1202 by Leonardo of Pisa, who is more commonly known as Fibonacci. The sequence starts like this: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34...
Each number is the sum of the two preceding ones. In addition to being fascinating, the Fibonacci sequence is found in nature and art. It appears in the arrangements of leaves on a stem, the breeding pattern of rabbits, and even in the spiral of galaxies. This natural appearance makes it a popular subject in both mathematical studies and cultural expressions.
In mathematical terms, the Fibonacci sequence is straightforward to understand yet holds many deep relationships and properties, leading to its widespread study and appreciation.

Closed Form Formula

The closed form formula offers an elegant way to calculate any Fibonacci number without having to compute all of the previous numbers in the sequence. The specific formula is: \[ F_n = \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{n+1} - \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n+1} \] When using this formula, you can directly determine the value of any Fibonacci number without iteration.
This is because the formula is derived from the characteristic equation of the linear recurrence relation that defines the Fibonacci sequence. The terms \( \frac{1+\sqrt{5}}{2} \) and \( \frac{1-\sqrt{5}}{2} \) are the roots of the equation, and are known as the golden ratio (\( \varphi \)) and its conjugate, respectively.
Their powers dominate the behavior of the sequence, which is why the closed form provides an exact computation for any \( n \). This formula combines the beauty of algebra and number theory.

Mathematical Induction

Mathematical induction is a powerful proof technique used to confirm that a statement is true for all natural numbers. The process involves two main steps:

• **Base Case:** Prove the statement for the first natural number (usually \( n = 0 \) or \( n = 1 \)).
• **Inductive Step:** Assume the statement holds for some arbitrary natural number \( k \), and then prove it holds for \( k+1 \).

This method is somewhat similar to dominoes falling. Once you show the first domino falls (the base case), demonstrating that any domino causes the following one to fall (inductive step) confirms all the dominoes will fall in sequence.
The beauty of mathematical induction lies in its simplicity and the depth of problems it can solve. It allows mathematicians to prove properties or formulas related to integers comprehensively, and it's widely used in various fields of mathematics and computer science.

Base Case

The base case is the first crucial step in a mathematical induction proof. It checks that the formula works for the initial number in your sequence or set.
In the context of proving the closed form for the Fibonacci sequence, we begin by verifying the base cases:

• \( F_0 \) using the formula results in 0.
• \( F_1 \) using the formula results in 1.

These values coincide with the definition of the Fibonacci sequence. Establishing the base case is essential because it provides the foundation upon which the entire inductive proof is built. Without a valid base case, the structure of the proof can't support further claims, as there's no starting point that confirms the formula's validity.
Therefore, always ensure to meticulously verify the base case when performing any proof by mathematical induction.