Using induction to prove formula I am revising for my test from Discrete math. I have come to this problem.

I am to prove by using mathematical induction that 

$6\times7^{n} - 2 \times 3^{n}$ is divisible by 4. for $n \ge 1$ ;

I created basic step :

$6\times7^{1} - 2\times3^{1} = 36 $ 

and induction step

$\forall n\ge 1, \exists K: 6\times7^{n} - 2\times3^{n} = 4K \Rightarrow  \forall n \ge 1, \exists l: 6\times7^{n+1} - 2\times3^{n+1} = 4l$

we can transform the formula into

$6\times7\times7^{n} - 2\times3\times3^{n}$

which is basicly

$42\times7^{n} -6\times3^{n}$

But what is the next step? I can i prove this fact?
HINT:

$$\begin{align*}
6\cdot7^{n+1}-2\cdot3^{n+1}&amp;=42\cdot7^n-6\cdot3^n\\
&amp;=7(6\cdot7^n)-3(2\cdot3^n)\\
&amp;=4(6\cdot7^n)+3(6\cdot7^n-2\cdot3^n)
\end{align*}$$
First, show that this is true for $n=1$:

$6\cdot7^{1}-2\cdot3^{1}=4\cdot9$

Second, assume that this is true for $n$:

$6\cdot7^{n}-2\cdot3^{n}=4\cdot{k}$

Third, prove that this is true for $n+1$:

$6\cdot7^{n+1}-2\cdot3^{n+1}=$

$6\cdot7\cdot7^{n}-2\cdot3\cdot3^{n}=$

$7\cdot6\cdot7^{n}-3\cdot2\cdot3^{n}=$

$7\cdot6\cdot7^{n}-(7-4)\cdot2\cdot3^{n}=$

$7\cdot6\cdot7^{n}-7\cdot2\cdot3^{n}+4\cdot2\cdot3^{n}=$

$7\cdot(\color\red{6\cdot7^{n}-2\cdot3^{n}})+4\cdot2\cdot3^{n}=$

$7\cdot\color\red{4\cdot{k}}+4\cdot2\cdot3^{n}=$

$4\cdot(7\cdot{k}+2\cdot3^{n})$



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