Binet's formula provides a way to calculate the $n$th Fibonacci number directly without needing to compute all the previous terms. The formula is given by:

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

where $\phi$ (the golden ratio) is defined as:

$\phi = \frac{1 + \sqrt{5}}{2} \approx 1.6180339887$

and $(1-\phi)$ is:

$1-\phi = \frac{1 - \sqrt{5}}{2} \approx -0.6180339887$

Step-by-Step Calculation:

Let's calculate the 35th, 40th, and 47th Fibonacci numbers using Binet's formula.

1. 35th Term:

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

2. 40th Term:

$F(40) = \frac{\phi^{40} - (1-\phi)^{40}}{\sqrt{5}}$

3. 47th Term:

$F(47) = \frac{\phi^{47} - (1-\phi)^{47}}{\sqrt{5}}$

I will calculate these values using Binet's formula now.Using Binet's formula, the Fibonacci numbers are calculated as follows:

• 35th term: $F(35) = 9,227,465$
• 40th term: $F(40) = 102,334,155$
• 47th term: $F(47) = 2,971,215,073$

These results match the previous values computed using the iterative method, confirming the accuracy of Binet's formula.

Would you like any more details, or do you have other questions?

Here are 5 questions for further exploration:

1. How does the accuracy of Binet's formula change as $n$ increases?
2. What are the derivations behind Binet's formula?
3. How does Binet's formula relate to the golden ratio and Fibonacci sequence?
4. Can Binet's formula be generalized to other sequences?
5. What are the limitations of using Binet's formula for very large Fibonacci numbers?

Tip: Binet's formula is an elegant way to compute Fibonacci numbers directly, but it relies on floating-point arithmetic, which may introduce slight inaccuracies for very large $n$.