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? 

For $n=k$, we have
$$12^k+2\cdot 5^{k-1} = 7M$$
Multiplying by $12$ throughout, we obtain
$$12^{k+1}+24\cdot 5^{k-1} = 84M \implies 12^{k+1} + (2\cdot 5 + 14)\cdot 5^{k-1} = 84 M$$
which in-turn implies
$$12^{k+1} + 2\cdot 5^k +14\cdot5^{k-1} = 84M \implies 12^{k+1}+2\cdot 5^k = 14\left(6M-5^{k-1}\right)$$
Hence, the statement is true for $n=k+1$.

We have
$$12^n = (7+5)^n = \sum_{k=0}^{n} \dbinom{n}k 7^k \cdot 5^{n-k} = 5^n + 7 \cdot \sum_{k=1}^{n} \dbinom{n}k 7^{k-1} \cdot 5^{n-k}$$
This gives us
\begin{align}
12^n + 2 \cdot 5^{n-1} & = 2\cdot 5^{n-1} + 5^n + 7 \cdot \sum_{k=1}^{n} \dbinom{n}k 7^{k-1} \cdot 5^{n-k}\\
& = \left(2+5\right)\cdot 5^{n-1} + 7 \cdot \sum_{k=1}^{n} \dbinom{n}k 7^{k-1} \cdot 5^{n-k}\\
& = 7 \cdot \left(5^{n-1} + \sum_{k=1}^{n} \dbinom{n}k 7^{k-1} \cdot 5^{n-k}\right)
\end{align}
