To solve $2F_{25} + F_{30}$ using Binet's formula, we first need to understand the general form of the Fibonacci sequence, denoted $F_n$, in terms of Binet's formula:

$F_n = \frac{1}{\sqrt{5}} \left( \alpha^n - \beta^n \right)$ where: $\alpha = \frac{1 + \sqrt{5}}{2}, \quad \beta = \frac{1 - \sqrt{5}}{2}$

Now, we can calculate $F_{25}$ and $F_{30}$, and then solve $2F_{25} + F_{30}$.

Step 1: Calculate $F_{25}$

$F_{25} = \frac{1}{\sqrt{5}} \left( \alpha^{25} - \beta^{25} \right)$

Step 2: Calculate $F_{30}$

$F_{30} = \frac{1}{\sqrt{5}} \left( \alpha^{30} - \beta^{30} \right)$

Step 3: Combine results for $2F_{25} + F_{30}$

The expression becomes: $2F_{25} + F_{30} = 2 \times \frac{1}{\sqrt{5}} \left( \alpha^{25} - \beta^{25} \right) + \frac{1}{\sqrt{5}} \left( \alpha^{30} - \beta^{30} \right)$ $= \frac{1}{\sqrt{5}} \left( 2\alpha^{25} - 2\beta^{25} + \alpha^{30} - \beta^{30} \right)$

Next, we compute the values of $\alpha^{25}, \beta^{25}, \alpha^{30}, \beta^{30}$, and substitute them back into the equation to obtain the result.

Let me calculate this for you.The value of $2F_{25} + F_{30}$ is approximately 982,090.

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:

1. How does Binet's formula approximate Fibonacci numbers for large $n$?
2. Why does the term $\beta^n$ become insignificant as $n$ increases?
3. Can Binet’s formula be used to calculate Fibonacci numbers in a computationally efficient way?
4. What is the relation between Fibonacci numbers and the golden ratio $\alpha$?
5. How does the Fibonacci sequence appear in nature and other mathematical contexts?

Tip: Binet's formula gives very accurate results for large Fibonacci numbers, but rounding errors can still occur due to floating-point precision limits.