Verify that the following sequence defined recursively admits as closed formula another sequence Verify that the following sequence defined recursively 
$\left \{ a_n \right \}_n \left \{ \begin{array}{rcl} a_0 & = & 0 & n = 0 \\ a_1 & = & 1 & n=1 \\ a_n & = & a_{n-1} + a_{n-2} & n \ge 2  \end{array} \right .$
admits as closed formula this sequence:
$\left \{ b_n \right \}_n = \frac{1}{\sqrt{5}} \cdot \left [ \left ( \frac{1 + \sqrt{5}}{2} \right )^n - \left ( \frac{1 - \sqrt{5}}{2} \right )^n \right ] \quad \forall n \ge 0$
I have verified the base step with $n=2$, i.e. I have to verify the following statement:
$P(2) : a_2 = b_2 \iff 1 = 1$  
but, I encounter some problems in the resolution of the inductive step:
I have set as true the statement:
$\begin{array}{lcl}P(n-2) : a_{n-2} & = & b_{n-2} \\ & = & \frac{1}{\sqrt{5}} \cdot \left [ \left ( \frac{1 + \sqrt{5}}{2} \right )^{n - 2} - \left ( \frac{1 - \sqrt{5}}{2} \right )^{n-2} \right ] \end{array}$
and from the truth of this last one I check the truth of 
$\begin{array}{lcl}P(n-1) : a_{n-1} & = & b_{n-1} \\ & = & \frac{1}{\sqrt{5}} \cdot \left [ \left ( \frac{1 + \sqrt{5}}{2} \right )^{n - 1} - \left ( \frac{1 - \sqrt{5}}{2} \right )^{n-1} \right ] \end{array}$
and from the truth of this last one I check the truth of 
$\begin{array}{lcl}P(n) : a_{n} & = & b_{n} \\ & = & \frac{1}{\sqrt{5}} \cdot \left [ \left ( \frac{1 + \sqrt{5}}{2} \right )^{n} - \left ( \frac{1 - \sqrt{5}}{2} \right )^{n} \right ] \end{array}$
therefore, 
$\begin{array}{lcl} a_{n} & = & a_{n-1} + a_{n-2} \\ & = & \frac{1}{\sqrt{5}} \cdot \left [ \left ( \frac{1 + \sqrt{5}}{2} \right )^{n-1} - \left ( \frac{1 - \sqrt{5}}{2} \right )^{n-1} \right ] + \frac{1}{\sqrt{5}} \cdot \left [ \left ( \frac{1 + \sqrt{5}}{2} \right )^{n-2} - \left ( \frac{1 - \sqrt{5}}{2} \right )^{n-2} \right ]\end{array}$
I have also to assign: $p = \frac{1+\sqrt{5}}{2}, q=\frac{1-\sqrt{5}}{2}$ obtaining
$\begin{array}{lcl} a_n & = & \frac{1}{\sqrt{5}} (p^{n-1} - q^{n-1} + p^{n-2} - q^{n-2}) \\ & = & \frac{1}{\sqrt{5}} ( p^np^{-1} - q^nq^{-1} + p^np^{-2} - q^nq^{-2}) \\ & = & \frac{1}{\sqrt{5}} [p^n(p^{-1} + p^{-2}) - q^n(q^{-1} + q^{-2})]\end{array}$  
I don't think if it is the right way, if I go to substitute appears a big expression. So this is my way and I stop here.  
Please, can you tell me how to resolve this exercise? Many Thanks!

Since $\frac{1+\sqrt{5}}{2}$ and $\frac{1-\sqrt{5}}{2}$ are roots of the polynomial $x^2-x-1=0$
You can substitute $p^2=p+1$ and $q^2=q+1$.
So
$$\frac{1}{\sqrt{5}}[p^{n-1}+p^{n-2}-q^{n-1}-q^{n-2}]=\frac{1}{\sqrt{5}}[p^{n-2}(p+1)-q^{n-2}(q+1)]=\frac{1}{\sqrt{5}}[p^n-q^n]$$
