To find $\frac{F_{35}}{5}$ using Binet's Formula, we can proceed as follows:

Binet's Formula for Fibonacci numbers:

$F_n = \frac{1}{\sqrt{5}} \left( \alpha^n - \beta^n \right)$ where:

• $\alpha = \frac{1 + \sqrt{5}}{2}$ (the golden ratio),
• $\beta = \frac{1 - \sqrt{5}}{2}$ (the conjugate of the golden ratio).

Now, let's calculate $F_{35}$ and then divide it by 5.

Step 1: Calculate $\alpha^{35}$ and $\beta^{35}$

• $\alpha^{35} = \left(\frac{1 + \sqrt{5}}{2}\right)^{35}$
• $\beta^{35} = \left(\frac{1 - \sqrt{5}}{2}\right)^{35}$

Since $\beta$ is a small negative number, $\beta^{35}$ will be very close to zero, so we can often approximate by ignoring $\beta^{35}$ for large $n$.

Step 2: Apply Binet's Formula

$F_{35} = \frac{1}{\sqrt{5}} \left( \alpha^{35} - \beta^{35} \right)$

We can approximate this, calculate $F_{35}$, and then divide by 5.

Let me calculate that.Using Binet's formula, we find that:

$\frac{F_{35}}{5} \approx 1,845,493$

Would you like more details on the steps, or have any further questions?

Here are 5 related questions to expand on this topic:

1. How is Binet's formula derived for the Fibonacci sequence?
2. Can we generalize this method for any $n$-th Fibonacci number?
3. How does $\beta^n$ affect the precision of Binet’s formula for large $n$?
4. What other methods exist for calculating Fibonacci numbers efficiently?
5. How is the golden ratio $\alpha$ connected to Fibonacci numbers?

Tip: For very large Fibonacci numbers, using approximations like Binet’s formula with careful handling of decimal precision can speed up the calculations.