#| This exercise has no tests.
Any solution is a right answer. |#


#| 
Step 1: Base cases
We start with the base cases of the Fibonacci sequence:
Fib(0) = 0
Fib(1) = 1

Step 2: Inductive hypothesis
Assume that for some k ≥ 1, Fib(k) = (φ^k - ψ^k)/√5, where ψ = (1 - √5)/2.

Step 3: Inductive step
We need to show that Fib(k+1) = (φ^(k+1) - ψ^(k+1))/√5.

Using the definition of the Fibonacci sequence, Fib(k+1) = Fib(k) + Fib(k-1).

Substituting the inductive hypothesis:
Fib(k+1) = (φ^k - ψ^k)/√5 + (φ^(k-1) - ψ^(k-1))/√5

Simplifying, we get:
Fib(k+1) = (φ^k + φ^(k-1) - ψ^k - ψ^(k-1))/√5

Next, we'll manipulate the expression to simplify it further.

Step 4: Manipulation
We know that φ and ψ are solutions to the quadratic equation x^2 = x + 1.

For φ, we have:
φ^2 = φ + 1

Rearranging, we get:
φ^2 - φ - 1 = 0

Similarly, for ψ, we have:
ψ^2 = ψ + 1

Rearranging, we get:
ψ^2 - ψ - 1 = 0

Now, let's express φ^k and ψ^k in terms of φ and ψ:
φ^k = φ^(k-1) * φ
ψ^k = ψ^(k-1) * ψ

Substituting these expressions into our previous equation, we get:
Fib(k+1) = (φ^k + φ^(k-1) - ψ^k - ψ^(k-1))/√5
= φ^(k-1) * (φ + 1) - ψ^(k-1) * (ψ + 1))/√5
= φ^(k-1) * φ^2 - ψ^(k-1) * ψ^2)/√5

Using the quadratic equations for φ and ψ, we can simplify further:
Fib(k+1) = φ^(k+1) - ψ^(k+1))/√5

Therefore, we have shown that Fib(k+1) = (φ^(k+1) - ψ^(k+1))/√5, assuming that Fib(k) = (φ^k - ψ^k)/√5.

Step 5: Conclusion
By induction, we have proven that for all n ≥ 0, Fib(n) = (φ^n - ψ^n)/√5.

Now, let's examine the expression (φ^n - ψ^n)/√5. We know that ψ < 1, and as n gets larger, ψ^n approaches 0. Therefore, the term ψ^n becomes negligible.

Thus, we can conclude that Fib(n) is the closest integer to φ^n/√5.
|#