Using Binets’ formula, solve for F50. (F sub 50

Binet's formula provides an explicit way to calculate the \( n \)-th Fibonacci number. The formula is given by:

\[
F_n = \frac{\phi^n - \psi^n}{\sqrt{5}}
\]

Where:
- \( \phi = \frac{1 + \sqrt{5}}{2} \) (the golden ratio)
- \( \psi = \frac{1 - \sqrt{5}}{2} \)
- \( n \) is the Fibonacci sequence position you're calculating.

For \( F_{50} \):
\[
F_{50} = \frac{\phi^{50} - \psi^{50}}{\sqrt{5}}
\]

Let's calculate this step-by-step.

1. Calculate \( \phi \) and \( \psi \):
   \[
   \phi = \frac{1 + \sqrt{5}}{2} \approx 1.6180339887
   \]
   \[
   \psi = \frac{1 - \sqrt{5}}{2} \approx -0.6180339887
   \]

2. Raise \( \phi \) and \( \psi \) to the 50th power:
   \[
   \phi^{50} \approx (1.6180339887)^{50}
   \]
   \[
   \psi^{50} \approx (-0.6180339887)^{50}
   \]

3. Subtract \( \psi^{50} \) from \( \phi^{50} \) and divide by \( \sqrt{5} \).

Calculating this directly using Binet's formula:

\[
F_{50} = \frac{(1.6180339887)^{50} - (-0.6180339887)^{50}}{\sqrt{5}}
\] 

Let me compute the value.Using Binet's formula, \( F_{50} \) is calculated to be:

\[
F_{50} = 12,586,269,025
\]

This is the 50th Fibonacci number.

Would you like a more detailed breakdown or have any questions?

Here are some related questions you might find interesting:
1. How is Binet's formula derived from the Fibonacci sequence?
2. What are some applications of Fibonacci numbers in real life?
3. How does the golden ratio (\( \phi \)) relate to Fibonacci numbers?
4. What is the significance of \( \psi \) in Binet's formula?
5. How can we approximate large Fibonacci numbers without using Binet's formula?

**Tip:** Binet's formula is an elegant way to calculate Fibonacci numbers directly without needing to generate all the preceding numbers, making it especially useful for large indices.

Learn how to calculate the 50th Fibonacci number using Binet's formula, which involves the golden ratio and the values of phi and psi. This step-by-step solution includes the application of algebraic functions and is suitable for students in Grades 9-12.

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