Proving that $12^n + 2(5^{n-1})$ is a multiple of 7 for $n\geq 1$ by induction 
Prove by induction that $12^n + 2(5^{n-1})$ is a multiple of $7$.

Here's where I am right now:
Assume $n= k $ is correct:
$$12^k+2(5^{k-1}) = 7k.$$
Let $n= k+1 $:
$$12^{k+1} + 2(5^k)$$
$$12^k(12) + 2(5^k)$$
Any ideas on how I can proceed from here? 

You should write $12^k+2\cdot 5^{k-1}=7r$ (using $k$ with two different meanings is wrong).
Then you have $12^k=7r-2\cdot 5^{k-1}$; now
\begin{align}
12^{k+1}+2\cdot 5^k
&=12\cdot 12^k+2\cdot 5^k\\
&=12\cdot(7r-2\cdot 5^{k-1})+2\cdot 5^k\\
&=7\cdot12r-24\cdot5^{k-1}+10\cdot5^{k-1}
\end{align}
and you should be able to conclude.
Remember to check the basis of the induction, that is, the case $n=1$.

Using congruences it is easier, of course: just observe that $12\equiv 5\pmod{7}$, so
$$
12^n+2\cdot 5^{n-1}\equiv 5^n+2\cdot 5^{n-1}\equiv 5^{n-1}(5+2)\equiv 5^{n-1}\cdot7\equiv0\pmod{7}
$$
