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:
$$F_N = \frac{1}{\sqrt5}((\frac{1+\sqrt5}{2})^N - (\frac{1-\sqrt5}{2})^N)$$
While the answer key states:

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.

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.
