Using mathematical induction prove that $ 11 \cdot 3^n + 3 \cdot 7^n - 6$ is divisible by 8 Prove that $ \phi(n) =11 \cdot 3^n + 3 \cdot 7^n - 6  $ is divisible by 8 for all $n \in N$.
Base: $ n = 0 $
$ 8 | 11 + 3 - 6 $ is obvious.
Now let $\phi(n)$ be true we now prove that is also true for $ \phi(n+1)$.
So we get $ 11 \cdot 3^{n+1} + 3 \cdot 7^{n+1} - 6$ and I am stuck here, just can't find the way to rewrite this expression so that I can use inductive hypothesis or to get that one part of this sum is divisible by 8 and just prove by one more induction that the other part is divisible by 8. 
For instance, in the last problem I had to prove that some expression a + b + c is divisible by 9. In inductive step b was divisible by 9 only thing I had to do is show that a + c is divisible by 9 and I did that with another induction, and I don't see if I can do the same thin here. 
 A: Suppose $11*3^n + 3*7^n - 6 = 8k$
The $11*3^{n+1} + 3*7^{n+1} - 6 = 11*3^n*3 + 3*7^n*7 - 6$
$=3(11*3^n + 3*7^n-2) + 4*3*7^n $
$= 3(11*3^n + 3*7^n - 6) + 4*3*7^n + 12$
$= 3(8k) + 4(3*7^n + 3)$;  $3*7^n$ is odd and $3$ is odd so $(3*7^n + 3)$ is even.
$= 3(8k) + 8(\frac{3*7^n + 3}2) = 8(3k +  \frac{3*7^n + 3}2)$.
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Actually I like and am inspired by Bill Dubuques answer.
We want to prove $\phi(n) = 11*3^n + 3*7^n - 6 \equiv 0 \mod 8$
And we know $\phi(n) = 11*3^n + 3*7^n - 6 \equiv 3*3^n + 3*(-1)^n -6 = 3^{n+1} + 3*(-1)^n - 6 \mod 8$.
So it's a matter of showing $f(n) = 3^{n+1} + 3(-1)^n \equiv 6 \mod 8$.
And if we notice $f(n+2) = 3^{n+3} + 3(-1)^{n+2} = 3^{n+1}*9 + 3(-1)^{n} \equiv 3^n + 3(-1)^{n}= f(n) \mod 8$.
So it's now just a matter of showing for $f(0) \equiv f(1) \equiv 6 \mod 8$.
Which is easily verified $3^1 + 3*(-1)^0 =3+3= 6$ and $3^2 + 3*(-1)^1 = 9 -3 = 6$
A: $\!\!\bmod 8\!: f(n\!+\!2)\equiv f(n)\,$ by $\,a\equiv 3,7\Rightarrow a^{\large 2}\!\equiv 1\Rightarrow a^{\large n+2}\!\equiv a^{\large n}.\,$ So $\,8\mid f(n)\!\!\iff\!\! 8\mid f(n\!+\!2)$ thus by (strong/parity) induction, it is true for all $n$ $\iff$ it is true for the base cases $\,n=0,1.$
A: Setup the same as your current work:
$\dots$
$\dots = 11\cdot 3^{n+1}+3\cdot 7^{n+1}-6 = 11\cdot 3\cdot 3^{n}+ 3\cdot 7\cdot 7^n - 6$
$=33\cdot 3^n + 21\cdot 7^n - 6 = (11+22)\cdot 3^n + (3 + 18)\cdot 7^n - 6$
$=\underbrace{11\cdot 3^n + 3\cdot 7^n - 6}_{\text{should be familiar}} + \underbrace{22\cdot 3^n + 18\cdot 7^n}_{\text{unknown}}$
Now, what can we say about $22\cdot 3^n+18\cdot 7^n$?  Anything?  You say in a previous example, you had to run a second induction proof to finish, might that be useful here?
A: Write $\phi(n) =11 \cdot 3^n + 3 \cdot 7^n - 6\cdot 1^n$. This is an integer linear combination of geometric sequences and so satisfies an integer linear recurrence. Indeed,
$$
\phi(n+3) = 11 \phi(n+2) - 31 \phi(n+1) + 21 \phi(n)
$$
The claim now follows from induction once you have proved the base cases for $n=0,1,2$.
The recurrence comes from the equation having $3$, $7$, and $1$ as roots: $$0=(x-3)(x-7)(x-1)=x^3 - 11 x^2 + 31 x - 21$$
If $u$ is root of that equation, then $u^{n+3} = 11 u^{n+2} - 31 u^{n+1} + 21 u^n$ for all $n$. For the argument above, the coefficients are irrelevant, except for the fact that they are integers.
