I need to prove the following equation using mathematical induction and using the phi values if necessary.
$\phi^n = \phi F_n + \text{(another Fibonacci number)}$
In this proof, it is kind of hard for me to prove this with the part "another Fibonacci number".
Thank you in advance.
Help is much appreciated.
It is easy to show $\phi^n = \phi F_n + F_{n-1}$ by induction.
We have $F_1=1,F_0=0$, so for $n=1$ we have to prove that $\phi=\phi+0$, which is true.
Suppose it is true for $n$.
By multiplying $\phi^n = \phi F_n + F_{n-1}$ (our supposition) with $\phi$ we then have $\phi^{n+1}=\phi^2 F_n+\phi F_{n-1}$.
Using $\phi^2=\phi+1$ the right hand side becomes $\phi F_n+F_n+\phi F_{n-1}=\phi(F_n+F_{n-1})+F_n=\phi F_{n+1}+F_n$, so the result is true for $n+1$.
Hint:
Assuming the conjecture true, compute $\phi^n-\phi F_n$ for increasing $n$:
$$\phi^1-\phi F_1=\phi-\phi=0,\\ \phi^2-\phi F_2=\phi^2-\phi=1,\\ \phi^3-\phi F_3=\phi^3-2\phi=\phi^3-\phi^2+\phi^2-2\phi=\phi^2-\phi=1,\\ \phi^4-\phi F_4=\phi^4-3\phi=\phi^4-\phi^3+\phi^3-\phi^2+\phi^2-3\phi=2\phi^2-2\phi=2,\\ \cdots $$
and the pattern seems to be
$$\phi^n=\phi F_n+F_{n-1}.$$
Then sum the equality for $n$ and $n+1$, giving
$$\phi^{n+1}+\phi^{n}=\phi^{n+2}=\phi(F_{n+1}+F_n)+(F_n+F_{n-1}).$$
You can conclude now.
I think this gets a bit simpler if you look at phi. It has a very special property, namely that
$$\phi^2 = \phi + 1 $$ Which you can generalize to n by induction since $$\phi^n = \phi^{n-1} * \phi $$
