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The problem asks you to prove by mathematical induction that for all integers \( n \geq 0 \), the Fibonacci sequence can be expressed as:

\[
F(n) = \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right)
\]

We will proceed with mathematical induction, following these steps:

### Step 1: **Base Case**

We need to verify the formula for the smallest value of \( n \), which is \( n = 0 \).

- The Fibonacci definition gives \( F(0) = 0 \).
- The formula for \( F(0) \) is:

\[
F(0) = \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^0 - \left( \frac{1 - \sqrt{5}}{2} \right)^0 \right)
\]

This simplifies to:

\[
F(0) = \frac{1}{\sqrt{5}} \left( 1 - 1 \right) = \frac{1}{\sqrt{5}} \cdot 0 = 0
\]

Thus, the base case holds true.

### Step 2: **Induction Hypothesis**

Assume that the formula holds for some arbitrary integer \( k \geq 0 \), i.e.,

\[
F(k) = \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^k - \left( \frac{1 - \sqrt{5}}{2} \right)^k \right)
\]

and

\[
F(k+1) = \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^{k+1} - \left( \frac{1 - \sqrt{5}}{2} \right)^{k+1} \right)
\]

### Step 3: **Induction Step**

We now need to show that if the formula holds for \( k \), then it also holds for \( k + 2 \), i.e.,

\[
F(k+2) = \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^{k+2} - \left( \frac{1 - \sqrt{5}}{2} \right)^{k+2} \right)
\]

By the definition of Fibonacci numbers:

\[
F(k+2) = F(k+1) + F(k)
\]

Using the induction hypothesis, we substitute \( F(k+1) \) and \( F(k) \) into the equation:

\[
F(k+2) = \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^{k+1} - \left( \frac{1 - \sqrt{5}}{2} \right)^{k+1} \right) + \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^k - \left( \frac{1 - \sqrt{5}}{2} \right)^k \right)
\]

Factor out \( \frac{1}{\sqrt{5}} \):

\[
F(k+2) = \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^k \left( \frac{1 + \sqrt{5}}{2} + 1 \right) - \left( \frac{1 - \sqrt{5}}{2} \right)^k \left( \frac{1 - \sqrt{5}}{2} + 1 \right) \right)
\]

Simplify the terms inside the parentheses:

\[
\frac{1 + \sqrt{5}}{2} + 1 = \frac{3 + \sqrt{5}}{2}
\quad \text{and} \quad
\frac{1 - \sqrt{5}}{2} + 1 = \frac{3 - \sqrt{5}}{2}
\]

We observe that:

\[
\frac{1 + \sqrt{5}}{2} \cdot \frac{3 + \sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2}^2
\quad \text{and} \quad
\frac{1 - \sqrt{5}}{2} \cdot \frac{3 - \sqrt{5}}{2} = \frac{1 - \sqrt{5}}{2}^2
\]

Thus, the expression simplifies to:

\[
F(k+2) = \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^{k+2} - \left( \frac{1 - \sqrt{5}}{2} \right)^{k+2} \right)
\]

Therefore, the induction step holds.

### Conclusion

By mathematical induction, we have proven that for all \( n \geq 0 \):

\[
F(n) = \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right)
\]

Would you like further details or clarifications?

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Here are some follow-up questions to extend your understanding:

1. Can you explain why the base case is crucial in mathematical induction?
2. How does the Fibonacci recurrence relation help simplify the proof in the induction step?
3. What role does the golden ratio play in the closed form of Fibonacci numbers?
4. Can you derive the closed form without using induction?
5. How does this proof relate to Binet's formula for Fibonacci numbers?

**Tip:** When proving statements by induction, always verify that your induction step aligns with the recurrence relation or formula structure of the sequence or problem!

Use mathematical induction to prove the Fibonacci closed-form formula F(n) = (1/√5) * (((1 + √5) / 2)^n - ((1 - √5) / 2)^n) for all integers n ≥ 0.

Prove the closed-form formula for Fibonacci numbers using mathematical induction. This problem explores algebraic sequences, the Fibonacci recurrence relation, and involves the golden ratio. Suitable for advanced high school or university-level students.

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