Let $a_n$ defined recursively by $a_0$ = 2, $a_1 = 4$ and
$a_{n+1}$ = 4$a_n$ - 3$a_{n-1}$, n $\ge$ 1
Solve the recurrence relation and verify the formula by using strong induction.
Attempt (not sure if it's right):
$\Rightarrow a_{n+1} - 4a_n + 3a_{n-1} = 0 \Rightarrow a^{n+1} - 4a^n + 3a^{n-1}$ = 0
$\Rightarrow a^2 - 4a + 3 = 0 \Rightarrow (a - 3)(a - 1) = 0 \Rightarrow a$ = 3,1
$\Rightarrow a_0 = 2 = B(3^n) + C = B + C$ = 2
$\Rightarrow a_1 = 4 = B(3^n) + C = 3B + C = 4$ $\Rightarrow B = 1, C = 1$
$\Rightarrow$ General formula: $a_n = 3^n + 1^n$
Hypothesis would be $ n > 1$ and $a_k \ge K$ for integers K such that $1 \le K < n$ ?
Not sure what to do next or if I was right up to this point
Edit (added another question)
$a_n = 5a_{n-1} - 6a_{n-2}, r_0=0$ and $r_1=1$
General formula $a_n= 3^n - 2^n$
Let us assume for some $n \ge 0, a_{n-1} = 3^{n-1} - 2^{n-1}$ and $a_{n-2} = 3^{n-2} - 2^{n-2}$ Is this the right hypothesis?
Thanks
Your approach is right, you did find the series. However, your redaction and reasoning are both terrible.
What you just wrote should not be on an answer. It is a draft that enables you to get the formula.
Once you have it, you can write the following.
$a_0=3^0+1$ and $a_1=3^1+1$. Let us assume that for some $n\geq 1$, $a_n=3^n+1$ and $a_{n-1}=3^{n-1}+1$. Let us calculate $a_{n+1}$
$a_{n+1} = 4(3^n+1)-3(3^{n-1}+1)=3^{n+1}+1$ (What a surprise! I didn't expect this at all...)
So by induction, $\forall n \in \mathbb{N}, a_n=3^n+1$
This kind of reasoning can be disturbing at first. But it is one valid way to proove your result. The disturbing fact is that the formula looks like it comes out of nowhere, you just tried formulas all day long and found one that fits. It is correct redaction, though.
Your implications do not prove anything. The first one (where you go from $a_n$ to $a^n$ somehow) is wrong, unless you assume some specific form for $a_n$ (and in this case you should change your notations).
Even if you had only valid implications to some property $A$, it would only show that any such sequence must verify property $A$, not that all sequences that verify property $A$ are solutions to your problem.