We are given

Fn=

1 Ö5

(an-bn)

where a and b are solutions of the quadratic equation x²-x-1=0 and a > b.

(1) In the quadratic equation ax²+bx+c=0 with roots a amd b, using Viete's relations for the sum and product of the roots, I obtain: ab = -b/a, a+ b = c/a
In the particular case of the equation x²-x-1=0, I have: ab = -1, a+ b = 1
(2)

1 a + 1 a2 = a+ 1 a2 =1

as a satisfies x²=x+1. Similarly for b.

(3) Here I shall prove that F₁=1, F₂=1 and F_n + F_n+1 = F_n+2 and hence F_n is the nth Fibonacci number. First, I calculate a and b:

a =

1+Ö5 2

and

a =

1+Ö5 2

F1 = 1 Ö5 æ ç è 1+Ö5 2 - 1-Ö5 2 ö ÷ ø = 1 F2 = 1 Ö5 (a2 - b2) = 1 Ö5 (a- b)(a+ b) = 1 Fn +Fn+1- Fn+2 = 1 Ö5 [an -bn + an+1 -bn+1 - an+2 + bn+2] Fn + Fn+1- Fn+2 = 1 Ö5 [-an(a2 - a- 1) +bn(b2- b-1)]=0

(4) I shall prove by induction the statement P_n that 1 + F₁ + F₂ + ¼F_n = F_n+2.

I know that F₁=1, F₂=1 and by (3), F₃=2, F₄=3, F₅=5, F₆=8,...
For n=1, P₁ is 1+ F₁ = F₃ which is evidently true as 1 + 1 = 2.
For n=2, P₁ is 1+ F₁ + F₂ = F₄ which is evidently true as 1 + 1 + 1 = 3.
For n=3, P₁ is 1+ F₁ + F₂ + F₃ = F₅ which is evidently true as 1 + 1 + 1 + 2 = 5.

Now I assume that P(k) is true for a fixed k and I shall prove that P(k+1) is also true, that is:

Pk : 1 + F1 + F2 + ¼Fk = Fk+2 Pk+1 : 1 + F1 + F2 + ¼Fk + Fk+1 = Fk+3

1 + F1 + F2 + ¼Fk + Fk+1 = (1 + F1 + F2 + ¼Fk ) + Fk+1 = Fk+2 + Fk+1 = Fk+3

and hence the result is true for n=k+1. By the principle of mathematical induction the statement P_n is true for all positive integer values of n which completes the proof.