Question

The Lucas sequence $L_n$ is defined by the recurrence relation: \[ L_n = L_{n-1} + L_{n-2}, \text{for } n \geq 3, \] with $L_1 = 1$ and $L_2 = 3$. Which one of the options given is TRUE?

• A. \(L_n = \left(\tfrac{1+\sqrt{5}}{2}\right)^n + \left(\tfrac{1-\sqrt{5}}{2}\right)^n\)
• B. \(L_n = \left(\tfrac{1+\sqrt{5}}{2}\right)^n - \left(\tfrac{1-\sqrt{5}}{3}\right)^n\)
• C. \(L_n = \left(\tfrac{1+\sqrt{5}}{2}\right)^n + \left(\tfrac{1-\sqrt{5}}{3}\right)^n\)
• D. \(L_n = \left(\tfrac{1+\sqrt{5}}{2}\right)^n - \left(\tfrac{1-\sqrt{5}}{2}\right)^n\)

Hint:

The Lucas sequence is closely related to the Fibonacci sequence. While Fibonacci numbers use $\frac{\alpha^n - \beta^n}{\alpha - \beta}$, the Lucas numbers use $\alpha^n + \beta^n$.

Answer 1

The Correct Option is A

Step 1: Characteristic equation.
The recurrence relation is: \[ L_n = L_{n-1} + L_{n-2} \] Its characteristic polynomial is: \[ x^2 - x - 1 = 0 \] Step 2: Solve for roots.
The roots are: \[ \alpha = \frac{1+\sqrt{5}}{2}, \beta = \frac{1-\sqrt{5}}{2} \] Step 3: General solution form.
The solution has the form: \[ L_n = A\alpha^n + B\beta^n \] Step 4: Use initial conditions.
For $n=1$: \[ L_1 = 1 = A\alpha + B\beta \] For $n=2$: \[ L_2 = 3 = A\alpha^2 + B\beta^2 \] Step 5: Known property of Lucas sequence.
The Lucas sequence is well-known to satisfy: \[ L_n = \alpha^n + \beta^n \] where $\alpha, \beta$ are the roots of $x^2 - x - 1 = 0$. 
Step 6: Verification.
Check $n=1$: \[ L_1 = \alpha + \beta = \frac{1+\sqrt{5}}{2} + \frac{1-\sqrt{5}}{2} = 1 \] Check $n=2$: \[ L_2 = \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 1^2 - 2(-1) = 3 \] This matches perfectly. \[ \boxed{L_n = \left(\frac{1+\sqrt{5}}{2}\right)^n + \left(\frac{1-\sqrt{5}}{2}\right)^n} \]