The tutor starts towards the general solution to a 2nd degree difference equation, aka recurrence relation.

Back in my June 21 post, the following recurrence relation emerged:

The tabulated relationship above can be written as t_n+2=t_n+1+t_n, and then as t_n+2 – t_n+1 – t_n = 0

The equation above is a second degree homogeneous recurrence equation. Its general solution will be of the form

t_n=as₁^n + bs₂^n,

where s₁, s₂ are the solutions to the related quadratic equation

x^2-x-1=0

Using the quadratic formula (see my post here), we get

x=(1+5^0.5)/2 or x=(1-5^0.5)/2

So the general solution to the recursive equation t_n+2 – t_n+1 – t_n = 0 will be

t_n = a((1+5^0.5)/2)^n + b((1-5^0.5)/2)^n

where a,b are constants to determine. Next post, I’ll show how to determine a and b in order to arrive at the definitive solution.

Source:

Grimaldi, Ralph P. Discrete and Combinatorial Mathematics. Don Mills:
   Addison-Wesley, 1994.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.