數學 - fibonacci ，MI（Prove）?

b)
By a) we have | Fn Fn+2 - F²n+1 | = | Fn-1 Fn+1 - F²n |
When n = 0 , |F0 F2 - F²1| = |1×2 - 1²| = |1| = 1 is true.
Assuming when n = k-1 , |Fk-1 Fk+1 - F²k| = 1 is true,
when n = k , by the result of a), | Fk Fk+2 - F²k+1 | = | Fk-1 Fk+1 - F²k | = 1 is true ,
by induction it is true for all n ≥ 0.

c)
By a) and b) we have (-1)(Fn-1 Fn+1 - F²n) = Fn Fn+2 - F²n+1 = ±1
When n = 1 , F0 F2 - F²1 = 1×2 - 1² = 1 = (-1)¹⁻¹ is true.
Assuming when n = k-1 , Fk-1 Fk+1 - F²k = (-1)ᵏ⁻¹ ⁻¹ is true,
when n = k , Fk Fk+2 - F²k+1 = (-1)(-1)ᵏ⁻¹ ⁻¹ = (-1)ᵏ⁻¹ is true.
By induction it is true for all n ≥ 1.

Missing question:

a) The pattern is F(n) = F(k) F(n-k) + F(k-1) F(n-(k+1))

Proof:
When k = 1 , F(n) = F(1) F(n-1) + F(1-1) F(n-2) = 1 F(n-1) + 1 F(n-2) is true.
Assuming when k = a , F(n) = F(a) F(n-a) + F(a-1) F(n-(a+1)) is true, then
F(n)
= F(a) F(n-a) + F(a-1) F(n-(a+1))
= F(a) ( F(n-(a+1)) + F(n-(a+2)) ) + F(a-1) F(n-(a+1))
= ( F(a) + F(a-1) ) ( F(n-(a+1)) + F(a) F(n-(a+2))
= F(a+1) F(n-(a+1)) + F(a) F(n-(a+2)) is true for k = a+1.

F0 = 0, F1 = 1, Fn = Fn-1 + Fn-2
(a) (Fn Fn+2 - F²n+1) + (Fn-1 Fn+1 - F²n) = 0

(b) Let P(n) : |Fn Fn+2 - F²n+1| = 1 for all n ≧ 0
When n = 0,
|F0 F2 - F²1|
= |(0)(1) - (1)²|
= |-1|
= 1
∴ P(0) is true.
Assume P(k) is true, i.e.
|Fk Fk+2 - F²k+1| = 1
When n = k + 1,
|Fk+1 Fk+3 - F²k+2|
= |Fk+1 (Fk+2 + Fk+1) - F²k+2|
= |Fk+1 Fk+2 + F²k+1 - F²k+2|
= |Fk+1 Fk+2 - F²k+2 + F²k+1|
= |Fk+2 (Fk+1 - Fk+2) + F²k+1|
= |Fk+2 (Fk+1 - Fk+1 - Fk) + F²k+1|
= |-Fk+2 Fk + F²k+1|
= |-(Fk Fk+2 - F²k+1|
= |-1|
= 1
∴ P(k+1) is also true.
By MI, P(n) is true for all n ≧ 0.

(c) Let Q(n) : Fn-1 Fn+1 - F²n = (-1)^n for all n ≧ 1
When n = 1,
F0 F2 - F²1
= (0)(1) - (1)²
= -1
= (-1)^(1)
∴ Q(1) is true.
Assume Q(k) is true, i.e.
Fk-1 Fk+1 - F²k = (-1)^k
When n = k + 1,
Fk Fk+2 - F²k+1
= -(Fk-1 Fk+1 - F²k) ⋯⋯⋯⋯ (from part (a))
= -(-1)^k
= (-1)^(k+1)
∴ Q(k+1) is also true.
By MI, Q(n) is true for all n ≧ 1.