# Help Solving $F_N = \frac{1}{\sqrt5}((\frac{1+\sqrt5}{2})^N - (\frac{1-\sqrt5}{2})^N)$ by Induction

I have recently got stuck on an induction problem in my textbook.

It is a big one so major kudos to anybody that can help me out.

The question states to prove this formula inductively: $$F_N = \frac{1}{\sqrt5}((\frac{1+\sqrt5}{2})^N - (\frac{1-\sqrt5}{2})^N)$$

The basis ( $N = 0$ and $N = 1$ ) is easily verified to be true. So assume that the theorem is true for all $0 ≤ i < N$ and we will establish for $N$. Let $φ_1 = (\frac{1+\sqrt5}{2})$ and $φ_2 = (\frac{1-\sqrt5}{2})$. Observe that both $φ_1$ and $φ_2$ satisfy $φ^N = φ^{N–1} + φ^{N–2}$ (this is verified by factoring and then solving a quadratic equation). Since $F_N = F_{N–1} + F_{N–2}$, by the inductive hypothesis we have

$F_N = \frac{1}{\sqrt5}\Bigl((φ_1)^{N-1} - ((φ_2)^{N-1}) + (φ_1)^{N-2} - (φ_2)^{N-2})\Bigr)$

= $\frac{1}{\sqrt5}\Bigl((φ_1)^{N} - ((φ_2)^{N})\Bigr)$

So far I have only managed to get up to the point where I get this: $$F_N = \frac{1}{\sqrt5}((\frac{1+\sqrt5}{2})^N - (\frac{1-\sqrt5}{2})^N) = \frac{1}{\sqrt5}\Bigl((φ_1)^{N-1} - ((φ_2)^{N-2})\Bigr)$$

After this I don't know what to do next. My questions are the following:

1) What actually is $F_N$? is it arbitrary or is it a sequence or something?

2) Do we verify $N=0$ and $N=1$ by plugging into this? $F_N = \frac{1}{\sqrt5}((\frac{1+\sqrt5}{2})^N - (\frac{1-\sqrt5}{2})^N) = \frac{1}{\sqrt5}\Bigl((φ_1)^{N-1} - ((φ_2)^{N-2})\Bigr)$

3) In the inequality $0\le i \lt N$ where did $i$ come from? What is $i$ used for?

4) How do we verify $φ^N = φ^{N–1} + φ^{N–2}$ by factoring and solving by quadratic equation?

5) lastly, what are we trying to find here? What does the question want as an answer? Usually I do proofs that have an expression = expression but this one is just fn = expression.

Any form of help or explanation would be greatly appreciated.

The link to the other thread about Binet's formula is full of overly complicated answers for a problem that is explained by a variant of a very, very simple identity, one that I have repeated time and time again: $$a^{n+1} - b^{n+1} = (a+b)(a^n - b^n) - ab(a^{n-1} - b^{n-1}).$$ This is trivially proven by expanding the RHS.

Now, let $$a = (1+\sqrt{5})/2, \quad b = (1-\sqrt{5})/2,$$ hence (again trivially) $$a+b = 1, \quad ab = \frac{1^2 - (\sqrt{5})^2}{2^2} = -1.$$ Therefore, we immediately establish that if $$F_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right),$$ we have $$F_{n+1} = F_n + F_{n-1}.$$ All that remains is to verify that $F_0$ and $F_1$ as such equals $0$ and $1$, respectively.

It is equally trivial to rephrase this proof in an inductive form, since the original identity is the inductive step. There is no need to appeal to the theory of difference equations, nor to generating functions. These underlying mechanisms yield insights into more general cases of linear recursions, but I am of the opinion that parsimony is not without its merits.

If you have a taste for linear algebra, try to follow/prove the following:

== Take the matrix $\;A:=\begin{pmatrix}1&1\\1&0\end{pmatrix}\in M_2(\Bbb R)\;$ , and observe that if we define $\;F_0=0,\,F_1=1,\,F_2=1,\,F_3=2,\,F_4=3,\ldots\;$ , then

$$A=\begin{pmatrix}F_2&F_1\\F_1&F_0\end{pmatrix}\;,\;\;A^2=\begin{pmatrix}F_3&F_2\\F_2&F_1\end{pmatrix}\;,\;\;\ldots$$

Prove Inductively: $$A^n=\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}\;\;\text{(this is very,very easy using the definition of Fibonacci sequence)}$$

== Now evaluate the characteristic polynomial of $\;A\;$ , and prove this matrix is diagonalizable over the reals. Use this to show that

$$A^n=P^{-1}\begin{pmatrix}\phi^n&0\\0&\phi^{-n}\end{pmatrix}P$$

with $\;P\;$ an invertible matrix (one formed with eigenvalues of $\;A\;$ ...) and $\;\phi=\frac{1+\sqrt5}2\;$ ...continue on.