Proving that $7^n(3n+1)-1$ is divisible by 9 I'm trying to prove the above result for all $n\geq1$ but after substituting in the inductive hypothesis, I end up with a result that is not quite obviously divisible by 9. 
Usually with these divisibility induction problems, it falls apart nicely and we can easily factorise say a 9 if the question required us to prove that the expression is divisible by 9. However in this case, I do not end up with such a thing.
My work so far below:
Inductive Hypothesis: $7^k(3k+1)-1=9N$ where $N\in\mathbb{N}$
Inductive Step:
$7^{k+1}(3k+4)-1 \\ =7\times 7^k(3k+1+3)-1 \\ =7\times \left [ 7^k(3k+1)+3\times 7^k \right ] -1 \\ = 7 \times \left [ 9N+1 + 3 \times 7^k \right ] -1 \\ = 63N+21\times 7^k+6 \\ = 3 \left [ 21N+7^{k+1}+2 \right ]$ 
So now I need to somehow prove that $21N+7^{k+1}+2$ is divisible by 3, but I'm not quite sure how to proceed from here...

First, show that this is true for $n=1$:
$7^1\cdot(3\cdot1+1)-1=9\cdot3$
Second, assume that this is true for $n$:
$7^n\cdot(3n+1)-1=9k$
Third, prove that this is true for $n+1$:
$7^{n+1}\cdot(3(n+1)+1)-1=$
$7^{n+1}\cdot(3n+3+1)-1=$
$7^{n+1}\cdot(3n+1+3)-1=$
$7^{n+1}\cdot(3n+1)+7^{n+1}\cdot(3)-1=$
$7\cdot\color{red}{7^n\cdot(3n+1)}+3\cdot7^{n+1}-1=$
$7\cdot(\color{red}{9k+1})+3\cdot7^{n+1}-1=$
$63k+7+3\cdot7^{n+1}-1=$
$63k+3\cdot7^{n+1}+6=$
$\color{blue}{3(21k+7^{n+1}+2)}$
Now:


*

*$7\equiv1\pmod3\implies$

*$\forall{m}\in\mathbb{N}:7^m\equiv1\pmod3\implies$

*$7^{n+1}\equiv1\pmod3\implies$

*$7^{n+1}+2\equiv3\pmod3\implies$

*$7^{n+1}+2\equiv0\pmod3\implies$

*$3|7^{n+1}+2\implies$

*$\exists{p}\in\mathbb{N}:7^{n+1}+2=3p\implies$

*$3(21k+7^{n+1}+2)=3(21k+3p)=3(3(7k+p))\color{blue}{=9(7k+p)}$



Please note that the assumption is used only in the part marked red.
