Ln = α^n + β^n proof?

Question

I need to show that Ln (the nth Lucas number) = α^n + β^n where α = (1+ √5)/2 and β = (1-√5)/2.

Answer 1

Tell your teacher that you don't need proof, you have faith!

Answer 2

Define Ln by L_0 = 2 L_1 = 1 L_2 = 3,
L_(n+2) = L_(n+1) + L_n
and the Fibonacci sequence by
F_0 = 0 F_1 = 1 F_2 = 1, F_(n+2) = F_(n+1) + F_n
The key to this problem is to show that
α^n = (L_n + F_n√5)/2
and β^n = (L_n - F_n√5)/2
Once you have those, then adding them gives your result.
I'll outline the proof of the first one. The second one is done
the same way.
We will proceed by induction on n.
The result is true for n = 1
and for n = 2, α² = (3+ √5)/2, so the result holds for n = 2.
So assume
α^n = (L_n + F_n√5)/2.
Let's try to verify the result for α^(n+1) and then we will be done.
α^(n+1) = α^n * α = (L_n + F_n√5)/2 *(1 + √5)/2
= (L_n + 5F_n + (L-n+F_n)√5)/4. (*)
But you can easily verify by induction that
L_n + 5F_n = 2*L_(n+1)
and
L_n + F_n = 2* F_(n+1).
(I'll let you fill in this step!)
and plugging this into (*) and dividing by 2
gives the result for α^(n+1).
This formula and the corresponding result for
F_n are called Binet's formulas.

Answer 3

This was done by Binet in 1843 i believe, perhaps you can look it up. Same idea for fibonacci sequence.