The Lucas sequence Ln is defined by the recurrence relation: Ln = Ln−1 + Ln−2,…
2023
The Lucas sequence Ln is defined by the recurrence relation:
Ln = Ln−1 + Ln−2, for n ≥ 3,
with L1 = 1 and L2 = 3.
Which one of the options given is TRUE?
- A.
\(L_n = \left(\frac{1 + \sqrt{5}}{2}\right)^n + \left(\frac{1 - \sqrt{5}}{2}\right)^n \) - B.
\(L_n = \left(\frac{1 + \sqrt{5}}{2}\right)^n - \left(\frac{1 - \sqrt{5}}{3}\right)^n \) - C.
\(L_n = \left(\frac{1 + \sqrt{5}}{2}\right)^n + \left(\frac{1 - \sqrt{5}}{3}\right)^n \) - D.
\(L_n = \left(\frac{1 + \sqrt{5}}{2}\right)^n - \left(\frac{1 - \sqrt{5}}{2}\right)^n \)
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Correct answer: A
We solve the recurrence L_n = L_{n-1} + L_{n-2} with initial values L1 = 1 and L2 = 3.
Characteristic equation: r^2 = r + 1
Find the roots: r = (1 + √5)/2 and r = (1 − √5)/2 (denote these by phi and psi).
General solution: L_n = A·phi^n + B·psi^n for constants A and B.
Use initial conditions: For n = 1: A·phi + B·psi = 1. For n = 2: A·phi^2 + B·psi^2 = 3.
Solve for A and B: Use identities phi + psi = 1 and phi·psi = −1. Compute phi^2 + psi^2 = (phi + psi)^2 − 2phi·psi = 1 − 2(−1) = 3.
Conclusion: Choosing A = 1 and B = 1 satisfies both initial equations, so L_n = phi^n + psi^n = ((1+√5)/2)^n + ((1−√5)/2)^n.
Therefore the correct closed form is Ln = ((1+√5)/2)^n + ((1−√5)/2)^n.