Binet's formula is used to find the $n$-th Fibonacci number without the need to compute all the preceding numbers. The Fibonacci sequence is defined as:

$F_n = F_{n-1} + F_{n-2}$ where $F_0 = 0$ and $F_1 = 1$.

Binet's formula is expressed as:

$F_n = \frac{\phi^n - \psi^n}{\sqrt{5}}$ where:

• $\phi = \frac{1 + \sqrt{5}}{2} \approx 1.61803$ (the golden ratio)
• $\psi = \frac{1 - \sqrt{5}}{2} \approx -0.61803$ (its conjugate)

Now, to calculate $F_{50}$, plug $n = 50$ into the formula:

$F_{50} = \frac{\phi^{50} - \psi^{50}}{\sqrt{5}}$

Let me calculate that for you.It seems like I can’t do more advanced data analysis right now. Please try again later. However, I can guide you on how to compute $F_{50}$ using Binet's formula with a calculator or any software that can handle large exponents.

1. Compute $\phi^{50}$, where $\phi = 1.61803$.
2. Compute $\psi^{50}$, where $\psi = -0.61803$.
3. Subtract $\psi^{50}$ from $\phi^{50}$.
4. Divide the result by $\sqrt{5}$, which is approximately 2.23607.

Do you want further details on any step? Here are some related questions:

1. How can I calculate powers of irrational numbers manually?
2. What is the significance of the golden ratio ($\phi$) in Fibonacci numbers?
3. How does Binet's formula approximate large Fibonacci numbers?
4. Can Binet’s formula be used for small Fibonacci numbers too?
5. How does the approximation of $\psi^n$ affect the accuracy of large Fibonacci numbers?

Tip: When using Binet's formula for large Fibonacci numbers, the contribution from $\psi^n$ becomes negligible, so $\phi^n$ alone gives a good approximation.