Math Problem Statement

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

Solution

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

Fn=Fn1+Fn2F_n = F_{n-1} + F_{n-2} where F0=0F_0 = 0 and F1=1F_1 = 1.

Binet's formula is expressed as:

Fn=ϕnψn5F_n = \frac{\phi^n - \psi^n}{\sqrt{5}} where:

  • ϕ=1+521.61803\phi = \frac{1 + \sqrt{5}}{2} \approx 1.61803 (the golden ratio)
  • ψ=1520.61803\psi = \frac{1 - \sqrt{5}}{2} \approx -0.61803 (its conjugate)

Now, to calculate F50F_{50}, plug n=50n = 50 into the formula:

F50=ϕ50ψ505F_{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 F50F_{50} using Binet's formula with a calculator or any software that can handle large exponents.

  1. Compute ϕ50\phi^{50}, where ϕ=1.61803\phi = 1.61803.
  2. Compute ψ50\psi^{50}, where ψ=0.61803\psi = -0.61803.
  3. Subtract ψ50\psi^{50} from ϕ50\phi^{50}.
  4. Divide the result by 5\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 (ϕ\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 ψn\psi^n affect the accuracy of large Fibonacci numbers?

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

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Math Problem Analysis

Mathematical Concepts

Fibonacci Sequence
Exponential Functions
Golden Ratio

Formulas

Binet's Formula: F_n = (phi^n - psi^n) / sqrt(5)
Golden Ratio: phi = (1 + sqrt(5)) / 2
Psi: psi = (1 - sqrt(5)) / 2

Theorems

Binet's Formula

Suitable Grade Level

Grades 10-12