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?
 A: HINT:
$$\begin{align*}
6\cdot7^{n+1}-2\cdot3^{n+1}&=42\cdot7^n-6\cdot3^n\\
&=7(6\cdot7^n)-3(2\cdot3^n)\\
&=4(6\cdot7^n)+3(6\cdot7^n-2\cdot3^n)
\end{align*}$$
A: You can add to $4K$ the number $36\cdot 7^n-4\cdot 3^n$ which clearly is divisible by 4 and you get that $4K+36\cdot 7^n-4\cdot 3^n=6\cdot 7^{n+1}-2\cdot 3^{n+1}$
A: If $6\cdot7^n-2\cdot3^n$ is divisible by 4, we can write it as $4k$ for some $k\in\Bbb N$.
Consider $24\cdot7^n$, which may be written as $4\cdot6\cdot7^n$ and is thus divisible by 4. Now add this to three times $6\cdot7^n-2\cdot3^n$:
$$24\cdot7^n+3(6\cdot7^n-2\cdot3^n)=42\cdot7^n-6\cdot3^n$$
$$=6\cdot7^{n+1}-2\cdot3^{n+1}=4(6\cdot7^n+3k)$$
So 4 divides $6\cdot7^{n+1}-2\cdot3^{n+1}$ if 4 divides $6\cdot7^n-2\cdot3^n$, and the inductive step is shown.
A: From your last step, you just need to replace $7^n$ by $\frac{4k+2.3^n}{6}$.
This gives $7[4k+2.3^n]-6.3^n=4[7k+2.3^n]$ which is divisible by $4$.
A: 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.
