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closed as a duplicate of: GATE CSE 2023 | Question: 5
The Lucas sequence $L_n$ is defined by the recurrence relation:
$L_n=L_{n-1}+L_{n-2}$, for $n \geq 3$ with $L_1=1$ and $L_2=3$.
Which one of the options given is TRUE?
  1. $L_n=\left(\frac{1+\sqrt{5}}{2}\right)^n+\left(\frac{1-\sqrt{5}}{3}\right)^n$
  2. $L_n=\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{3}\right)^n$
  3. $L_n=\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n$
  4. $L_n=\left(\frac{1+\sqrt{5}}{2}\right)^n+\left(\frac{1-\sqrt{5}}{2}\right)^n$
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The given recurrence relation is $L_n = L_{n-1} + L_{n-2}$

$\therefore L_n - L_{n-1} - L_{n-2} = 0$

Therefore, the characteristic equation is $r^2 - r -1 = 0$

Solving for $r$ gives $r_1 = \frac{1+\sqrt5}{2}, r_2 = \frac{1-\sqrt5}{2}$

Therefore, the general solution is of the form $L_n = \alpha(r_1)^n + \beta(r_2)^n$

$\therefore L_n = \alpha(\frac{1+\sqrt5}{2})^n + \beta(\frac{1-\sqrt5}{2})^n$

Now, given $L_1 = 1$

$\therefore L_1 = 1 = \alpha(\frac{1+\sqrt5}{2}) + \beta(\frac{1-\sqrt5}{2}) = (\alpha + \beta) \frac{1}{2} + (\alpha - \beta) \frac{\sqrt5}{2}$

$\therefore \alpha + \beta = 2 \text{ and } \alpha - \beta = 0$

$\therefore \alpha = 1, \beta = 1$

$\therefore L_n = (\frac{1+\sqrt5}{2})^n + (\frac{1-\sqrt5}{2})^n$

Answer :- D.

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