Prove that $2^n +1$ is divisible by $3$ for all positive integers $n$. I just want to know if I went on the right direction. With induction
Let $n=1$, then $2^1+1= 3$, which is divisible by $3$. Then show proof for 
$n+1.$
$2^n+1=3k$
So we get $2^{n+1}+1, \rightarrow 2^n+2+1, \rightarrow 3k+3= 3(k+1)$. Thus $2^n+1$ is divisible by $3$.
Now if I wanted to show that $2^n+1$ is divisble by $3$, $\forall$ odd integers $n$.
Would it be with induction:
$n=1$, then $2+1=3$, and $3|3$.
Let $n=2k+1$, since n is odd, then we get $2^{2k+1}+1=3m$. Now we need to show for $k+1$. 
We get: $2^{2k+2+1}+1=3m \rightarrow 2^{2k+1}*2^2+4-3 \rightarrow 4(2^{2k+1}+1)-3$
$\rightarrow 4(3m)-3 \rightarrow 3(4m-1)$, thus $2^n+1$ is divisible by $3$.

If you need to prove it for odd $n$, then here is an elegant proof
$$2^n=(3-1)^n$$
$$=3k+(-1)^n$$
Since $n$ is odd , $(-1)^n=-1$
Also proving $2^n+1$ is divisible by $3$ is same as proving $2^n-2$ is divisible by $3$
Therefore
$$ 2^n-2= (3-1)^n-2$$
$$=3k+(-1)^n-2$$
$$=3k-1-2$$
$$3k-3 $$
And this is divisible by $3$ for all $n\in \mathbb N(odd)$ 
And of course it is not true for every even $n$
