discrete mathematics - Solving Strong Mathematical Induction Sequence

I'm trying to work on the problem below, though I've hit a wall on how to proceed to prove the inductive step.

Suppose that $c_0,c_1,c_2\ldots$ is a sequence defined as follows: $$c_0=2,\, c_1=2,\, c_2=6,\, c_k=3c_{k-3}\,(k\geq3)$$

Prove that $c_n$ is even for each integer $n\geq0$.

Here is what I have so far:

1. Show that $P(0)$ and $P(1)$ are true.

   
   
   
   
   
   
   • $c_0=2$ and $2\geq0$ and $2\mid2$, so this is even.
   
   
   • $c_1=2$ and $2\geq0$ and $2\mid2$, so this is even.
   
   



2. Show that for every integer $k\geq1$, if $P(i)$ is true for each integer $i$ with $0\leq i\leq k$, then $P(k+1)$ is true.

   
   
   
   
   
   • Let $k$ be any integer with $k\geq1$, and suppose $c_i$ is even for each integer $i$ with $0\leq i\leq k$ [inductive hypothesis].
   
   
   • I must show that $c_{k+1}$ is even for each integer $k\geq0$.
   
   
   • Now $c_{k+1}=3c_{k-2}$...
   
   
   

...and this is where I do not understand how to proceed. Any tips on how I can finish solving this problem are greatly appreciated.

Answer

Let $P(n)$ be the statement “$c_n$, $c_{n+1}$, $c_{n+2}$ are even”. We have $P(0)$ by assumption. Furthermore, $P(n)\Rightarrow P(n+1)$, since if $c_n$, $c_{n+1}$, $c_{n+2}$ are even, $c_{n+3}=3c_n$ must be even also. By induction, $P(n)$ is true for all $n$, which implies immediately that $c_n$ must be even for all $n$.

A more immediate solution would be to use @fleablood’s approach, but this one enables you to do this problem by pure, vanilla induction.