GATE CS 2023 Q.15: Combinatorics, Counting, Recurrence Relations and Generating Functions

The Lucas sequence $L_n$ is defined by the recurrence relation:

L_n = L_{n-1} + L_{n-2}, \quad \text{for } n \ge 3,

with $L_1 = 1$ and $L_2 = 3$ .

Which one of the options given is TRUE?

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

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

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

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

Official answer and solution

Official answer: A
From: Official GATE 2023 answer key

Solution

Solve the second-order linear homogeneous recurrence with the characteristic equation.

The recurrence is $L_n - L_{n-1} - L_{n-2} = 0$ . Its characteristic equation is $r^2 - r - 1 = 0$ , having roots:

r_1 = \frac{1 + \sqrt{5}}{2}, \quad r_2 = \frac{1 - \sqrt{5}}{2}

The general solution is $L_n = c_1 r_1^n + c_2 r_2^n$ .

Using the initial conditions:

For $n = 1$ : $c_1 r_1 + c_2 r_2 = 1$
For $n = 2$ : $c_1 r_1^2 + c_2 r_2^2 = 3$

Notice that $r_1 + r_2 = 1$ and $r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2 r_1 r_2 = 1^2 - 2(-1) = 3$ . Hence, $c_1 = 1$ and $c_2 = 1$ satisfy both initial conditions directly.

Therefore, $L_n = \left(\frac{1 + \sqrt{5}}{2}\right)^n + \left(\frac{1 - \sqrt{5}}{2}\right)^n$ .

Final answer

Option (A)

Common trap

Confusing Lucas numbers with Fibonacci numbers (which have a $1/\sqrt{5}$ factor and a minus sign).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2023 Computer Science and Information Technology (CS)
Question: 15 of 65
Section: Computer Science and Information Technology

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