This sounds confusing, but I will try to make it easy. The question is to prove the proposition
$P(n): 3^n – 1$ is divisible by $2$.
So I first used $k=1$, and saw that it works out to be true.
Then when I started to work on part 2, I got lost quickly. So I stated that we assume true for $P(k)$ which means $3^k - 1$ is divisible by 2.$
I then used $P(K)$ to try and prove, $3^{k+1} -1 = 3 \cdot 3^k -1$ is divisible by 2.
I am not sure what to do after this step since I'm accustom to proving propositions in the form of equations, and here, there is not another side of any equation.
Does the proof just end there?
proposition: $2$ divides $3^n - 1$
Proof by induction.
Base case: $n=1, 2$ divides $4$
Inductive hypothesis: Suppose for some $n, 2$ divides $3^n - 1$
We will show that: $2$ divides $3^{n+1} - 1$ when $2$ divides $3^n - 1$
$3^{n+1} - 1 = 3(3^n-1)+2$
$2$ divides $(3^n-1)$ from the inductive hypothesis, and $2$ divides $2$
QED