Prove by induction $6\cdot 7^n -2\cdot 3^n$ is divisible by $4$ for $n\ge 1$ Prove by induction $6\cdot 7^n -2\cdot 3^n$ is divisible by $4$ for $n\ge 1$.
I am struggling on this problem very much.
So far I have Basis case = $6\cdot 7^1 - 2\cdot3^1 = 36$ which is divisible by $4$
Assume for a $n$ that $6\cdot 7^n-2\cdot3^n$ is divisible by $4$.
After that I am not sure what to do.
 A: $$6\cdot7^{n+1}-2\cdot3^{n+1}=7(6\cdot7^n-2\cdot3^n)+14 \cdot 3^n-2 \cdot 3^{n+1}=7(6\cdot7^n-2\cdot3^n)+8 \cdot 3^n$$
The first term is divisible by 4  by  inductive assumption and $8= 4 \cdot 2$
A: If $a$ is divisible by $4$, then $b$ is divisible by $4$ if and only if $b-a$ is divisible by $4$.
So, you can assume $6\cdot 7^n-2\cdot 3^n$ is divisible by $4$ and try for
\begin{align}
(6\cdot 7^{n+1}-2\cdot 3^{n+1})-(6\cdot 7^n-2\cdot 3^n)
&=
(6\cdot 7^{n+1}-6\cdot 7^{n})-(2\cdot 3^{n+1}-2\cdot 3^{n})
\\
&=
(6\cdot 7\cdot 7^{n}-6\cdot 7^{n})-(2\cdot 3\cdot 3^{n}-2\cdot 3^{n})\\
&=
\dots
\end{align}

Of course, using congruences modulo $4$ is easier:
$$
6\cdot 7^n-2\cdot 3^n\equiv 2\cdot 3^n-2\cdot 3^n\equiv 0 \pmod{4}
$$
A: Let $P(n) = 6\cdot 7^n - 2\cdot 3^n$. If $4| P(1)$ and we assume $P(k)$ to be true for some arbitrary natural number $k$ and we can show $4| P(k+1) - P(k)$ then we have proved that $4|P(n) \ \forall n\in\mathbb{N}$.
$$\begin{align*} P(k+1) - P(k) & =  6\cdot 7^{k+1} - 2\cdot 3^{k+1} - 6\cdot 7^k + 2\cdot 3^k \\ & = 7\cdot 6\cdot 7^k - 3\cdot 2\cdot 3^k - 6 \cdot 7^k + 2\cdot 3^k \\ & = 6\cdot 6 \cdot 7^k - 2 \cdot 2 \cdot 3^k \\ & = 4 \cdot (9\cdot 7^k - 3^k)\end{align*}$$
A: $\ f_{n} = ca^n\Rightarrow\, f_{n+1}\! - f_{n}\, =\, ca^{n+1}\!-ca^n = (\color{#c00}{a\!-\!1}) ca^n$  
Applying this to  $\,f_{n} = 6\cdot 7^n\!-2\cdot 3^n$ yields  $\color{#0a0}{f_{n+1}-f_{n}}\, =\ \smash[t]{\overbrace{\color{#c00}6\cdot 6}^{\large 4\cdot 9}\cdot 7^n -\overbrace{\color{#c00}2\cdot 2}^{\large 4\cdot 1}\cdot 3^n}$ is a multiple of $\,4,\,$ hence so too is $\,\color{#0a0}{f_{n+1}-f_n} + f_n = f_{n+1},\,$ since $\,4\mid f_n\,$ by induction. That is the inductive step.
Remark $\ $ This can be viewed as a special case of telescopic induction, a powerful method that serves to unify and simplify such forms of induction. Namely, telescopy shows that $\,f_n\,$ is the sum of its differences $\,\color{#0a0}{f_{n}-f_{n-1}}$ which, by above, are multiples of $4,\,$ hence so to is $\,f_n.$ Thus telescopy reduces the induction to the trivial induction that a sum of multiples of $4$ remains a multiple of $4$. In the same way, telescopy trivializes many inductive proofs involving sums and products, and lends great insight into the essence of such inductive proofs. You can find many further examples in my prior posts on telescopy.
A: Of course induction is a powerful tool, but I am not sure it should be the starting point, when a direct answer is possible. This would otherwise in the end "force" the student into rushing towards induction whenever n appears in a problem and prevent him from looking at more (and simpler angles)
In this case it is very simple $6.7^n -2.3^n = 2.(3.7^n-3^n)$
both elements inside the parenthesis are the products of odd numbers (3 and n 7s for the first one and n 3s for the second one), so they are both odd and thus their difference is even, i.e. A multiple of 2. We already have a 2 before the parenthesis, so the total is a multiple of 4. 
