Proving a Fibonacci Sequence

Question

I am currently trying to prove a Fibonacci sequence for my class and I am stuck.

$$F(n+1)^2 =F(n)^2 + F(n−1)F(n+2)\quad \text{for}\quad n\ge 2$$

What I've done so far:

$F(n+2) = F(n+1)+F(n)$

$\Rightarrow F(n+1) = F(n+2)-F(n)$

$[F(n+2)-F(n)]^2 = F(n+2)^2 - 2[F(n+2)F(n)] + F(n)^2$

$\dots\,$?

I am not quite sure I'm doing this correct or I just need to keep working at it. Any help would be appreciated.

Answer 1

At this point we want to show that $$F(n+2)^2-2F(n+2)F(n)=F(n-1)F(n+2)$$ in order to complete the proof. Well we have that $$\begin{aligned} F(n) &= F(n+1)-F(n-1) \\ &= F(n+2)-F(n)-F(n-1) \end{aligned}$$ by simply using the fact that $F(k+1)=F(k+2)-F(k)$ twice. Solving for $F(n)$ yields $F(n)=\frac{1}{2}[F(n+2)-F(n-1)]$. Try using this.

Answer 2

Hint. For all $(a,b)\in\mathbb{R}^2$, one has: $$a^2-b^2=(a-b)(a+b).$$ Apply this identity to $a=F_{n+1}$ and $b=F_n$.

Answer 3

Given that $F(n+1) = F(n) + F(n-1)$ you have

$F(n+1)^2 = F(n)^2 + 2F(n)F(n-1) + F(n-1)^2 = $

$F(n)^2 + F(n-1)[2F(n) + F(n-1)] = $

$F(n)^2 + F(n-1)[F(n) + (F(n)+F(n-1))] = $

$F(n)^2 + F(n-1)[F(n) + F(n+1)] = $

$F(n)^2 + F(n-1)F(n+2)$ QED

Answer 4

The proofs already provided are elegant. Here is another method. If we use characteristic equations, then the equation for $F_n$ is $$x^2=x+1,$$ and for any of $F_{n+1}^2$, $F_n^2$, $F_{n-1}F_{n+2}$, it is $$x^3=2x^2+4x+1,$$ so is their linear combination $$G_n = F_{n+1}^2-F_{n}^2-F_{n-1}F_{n+2}.$$

In other words, we have $$G_{n+3}=2G_{n+2}+4G_{n+1}+G_n$$ for $n\ge 2$.

On the other hand, we can verify that the $G_2=G_3=G_4=0$, so $$G_n=0$$ for all $n\ge 2$.

(The roots of the characteristic equation of $F_n$ are $\alpha=\frac{1+\sqrt{5}}{2}$ and $\beta=\frac{1-\sqrt{5}}{2}$. The roots of the characteristic equation for any homogeneous quadratic form of $F_n$ are $\alpha^2$, $\beta^2$, and $\alpha\beta=-1$. )