<s>[INST] 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?
 [/INST] 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.
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