Binet's formula for the Fibonacci sequence is given by:

$F(n) = \frac{\phi^n - (1 - \phi)^n}{\sqrt{5}}$

where:

• $F(n)$ is the $n$-th Fibonacci number,
• $\phi = \frac{1 + \sqrt{5}}{2}$ is the golden ratio (approximately 1.618),
• $1 - \phi = \frac{1 - \sqrt{5}}{2}$ is approximately -0.618.

Now let's calculate the 15th, 25th, and 35th Fibonacci numbers using this formula.

1. 15th Fibonacci number ($F(15)$):

$F(15) = \frac{\phi^{15} - (1 - \phi)^{15}}{\sqrt{5}}$

2. 25th Fibonacci number ($F(25)$):

$F(25) = \frac{\phi^{25} - (1 - \phi)^{25}}{\sqrt{5}}$

3. 35th Fibonacci number ($F(35)$):

$F(35) = \frac{\phi^{35} - (1 - \phi)^{35}}{\sqrt{5}}$

Let me calculate these values for you using Binet's formula.Using Binet's formula, the Fibonacci numbers are:

• $F(15) = 610$
• $F(25) = 75,025$
• $F(35) = 9,227,465$

Let me know if you'd like further details or clarifications.

Here are five related questions to deepen your understanding:

1. What are the first 10 Fibonacci numbers using Binet’s formula?
2. How does Binet’s formula work for large Fibonacci numbers, say for $F(100)$?
3. Can you prove Binet’s formula using the properties of the Fibonacci sequence?
4. What is the relationship between the golden ratio and the Fibonacci sequence?
5. Can you explain how the error term behaves in Binet's approximation for small $n$?

Tip: Binet’s formula provides an exact solution but relies on floating-point approximations, so it's best suited for moderate-sized Fibonacci numbers.