Skip to content

GATE CS 2023 Q5 — Combinatorics: Counting, Recurrence Relations, Generating Functions

Explain with AI

The Lucas sequence LnL_{n} is defined by the recurrence relation:
Ln=Ln−1+Ln−2, for n≥3,L_{n}=L_{n-1}+L_{n-2}, \quad \text{ for } \quad n \geq 3,
with L1=1L_{1}=1 and L2=3L_{2}=3.

Which one of the options given is TRUE?

Choose one option
View worked solution

Answer: A

The recurrence relation is given by Ln=Ln−1+Ln−2L_n = L_{n-1} + L_{n-2} for n≥3n \geq 3 with L1=1L_1 = 1 and L2=3L_2 = 3.

The characteristic equation of this recurrence relation is:

r2−r−1=0r^2 - r - 1 = 0

The roots of this equation are:

r1=1+52andr2=1−52r_1 = \frac{1+\sqrt{5}}{2} \quad \text{and} \quad r_2 = \frac{1-\sqrt{5}}{2}

The general solution is of the form Ln=c1r1n+c2r2nL_n = c_1 r_1^n + c_2 r_2^n. We can find the correct option by testing the base cases n=1n=1 and n=2n=2 for Option A:

  • For n=1n=1: L1=(1+52)1+(1−52)1=1+5+1−52=1L_1 = \left(\frac{1+\sqrt{5}}{2}\right)^1 + \left(\frac{1-\sqrt{5}}{2}\right)^1 = \frac{1+\sqrt{5} + 1-\sqrt{5}}{2} = 1
  • For n=2n=2: L2=(1+52)2+(1−52)2=6+254+6−254=124=3L_2 = \left(\frac{1+\sqrt{5}}{2}\right)^2 + \left(\frac{1-\sqrt{5}}{2}\right)^2 = \frac{6+2\sqrt{5}}{4} + \frac{6-2\sqrt{5}}{4} = \frac{12}{4} = 3

Since both initial conditions are satisfied, Option A is the correct closed-form expression.

Free practice · Your answers on this page are not saved.

Published
Report a correction
Keep exploringMore Combinatorics: Counting, Recurrence Relations, Generating Functions questions