The tutor completes the general solution to the 2nd degree recurrence relation.

In my previous post I began seeking the general solution to the recurrence relation tabulated below:

Specifically, I claimed its solution to be of the form

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

We need to find a and b such that the tabulated values for t_n come true. First of all, for n=0, we have

t₀ = a((1+5^0.5)/2)^0 + b((1-5^0.5)/2)^0

which simplifies to

1=a+b

Next, for n=1, we have

t₁=1=a((1+5^0.5)/2)^1 + b((1-5^0.5)/2)^1

which becomes, with b=1-a,

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

Multiplying both sides by 2, we get

2=a(1+5^0.5)+(1-a)(1-5^0.5)

which leads to

2=a+a5^0.5+1-5^0.5-a+a5^0.5

and then

2=2a5^0.5+1-5^0.5

Subtracting 1 then adding 5^0.5 to both sides gives

5^0.5+1=2a5^0.5

Next, dividing both sides by 2(5^0.5), we get

(5^0.5 +1)/(2(5^0.5))=a

Recalling that b=1-a, we have

b=(5^0.5 -1)/(2(5^0.5))

Believe it or not, the solution to the tabulated recurrence relation is

t_n = (5^0.5 +1)/(2(5^0.5))((1+5^0.5)/2)^n + (5^0.5 -1)/(2(5^0.5))((1-5^0.5)/2)^n
Source:

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

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