So I'm going over some examples of recursion and Fibonacci Sequences for my quiz tomorrow and I'm a bit lost after a certain point.
Prove $f_{n}^{2} = f_{n-1}f_{n+1}+(-1)^{n+1}$ $n\geq 2$
Basis Step: P(2)
$f_{2}^{2} = f_{2-1}f_{2+1}+(-1)^{n+1}$
$(1)^{2} = f_{1}f_{3}+(-1)^{3}$
$(1)^{2} = (1)(2)+(-1)$
$1 = 1$
Induction Step: Assume P(n) is true, Prove P(n+1), that is, show: $f_{n+1}^{2} = f_{n}(f_{n}+f_{n+1})+(-1)^{(n+2)}$
$$\begin{align} f_{n}f_{n+2}+(-1)^{n+2} & = f_{n}(f_{n+1}+f_{n})+(-1)^{n+2} \\ & = f_{n}f_{n+1}+f_{n}^{2}+(-1)^{n+2} \\ & = f_{n}f_{n+1}+[f_{n-1}f_{n+1}+(-1)^{n+1}]+(-1)^{n+2} \\ \end{align}$$
What do I do after this step? I'm a bit lost...
In your inductive step you started out by assuming $$f_{n+1}^2=f_nf_{n+2}+(-1)^{n+2}\ .$$ But this is what you are trying to prove, so your argument, even if "successful", will be circular and therefore wrong.
A better way would be to start with the RHS only of this equation and prove it equals the LHS. You could begin $$f_nf_{n+2}+(-1)^{n+2}=f_n(f_{n+1}+f_n)+(-1)^{n+2}\ .$$ See if you can take it from here. Remember that at some stage you will have to use the assumed formula for $f_n^2$.