Gate CS 2023 | Question - 15 | Discrete Mathematics

The Lucas sequence L_n is defined by the recurrence relation:
L_n = L_n−1 + L_n−2, for n ≥ 3,

with L₁ = 1 and L₂ = 3.

Which one of the options given is TRUE?

$L_{n} = {(\frac{1 + \sqrt{5}}{2})}^{n} + {(\frac{1 - \sqrt{5}}{2})}^{n}$

$L_{n} = {(\frac{1 + \sqrt{5}}{2})}^{n} - {(\frac{1 - \sqrt{5}}{3})}^{n}$

$L_{n} = {(\frac{1 + \sqrt{5}}{2})}^{n} + {(\frac{1 - \sqrt{5}}{3})}^{n}$

$L_{n} = {(\frac{1 + \sqrt{5}}{2})}^{n} - {(\frac{1 - \sqrt{5}}{2})}^{n}$

Solution:

L_n = L_n−1 + L_n−2 (for n ≥ 3_

L₁ = 1 and L₂ = 3

Put n=1 in option A

$L_{1} = {(\frac{1 + \sqrt{5}}{2})}^{1} + {(\frac{1 - \sqrt{5}}{2})}^{1}$

L₁ = 1

Put n=2 in option A

$L_{2} = {(\frac{1 + \sqrt{5}}{2})}^{2} + {(\frac{1 - \sqrt{5}}{2})}^{2}$

$L_{2} = \frac{{(1 + \sqrt{5})}^{2} + {(1 - \sqrt{5})}^{2}}{4}$

As we know, (a + b)² + (a - b)² = 2(a² + b²)

$L_{2} = \frac{2 (1 + 5)}{4}$

L₂ = 3

Option A satisfies the condition, so the correct answer is Option A.

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