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.

Learn how to use Binet's formula to calculate F35/5, including the application of the golden ratio and approximations for large Fibonacci numbers. This problem provides a detailed explanation suitable for high school students in Grades 10-12.

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