<s>[INST] 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? 
 [/INST] Basis Step:
$P(1) = 12^1 + 2\cdot5^{1-1} = 12 + 2 = 14 = 7\cdot 2$
Thus $P(1)$ is true.
Inductive Step:
Assume $P(k)$ holds for an arbitrary positive integer $k$. Under this assumption, let us prove that $P(k+1)$ is true, namely that
$$12^{k+1} + 2\cdot5^{k}  $$ is also a multiple of 7.
\begin{align}
12^{k+1}+2\cdot 5^k
&=12\cdot 12^k+2\cdot 5^k\\
&=12\cdot(7q-2\cdot 5^{k-1})+2\cdot 5^k\\
&=7\cdot12q-24\cdot5^{k-1}+10\cdot5^{k-1}\\
&=7\cdot12q+(-24+10)\cdot5^{k-1}\\
&=7\cdot12q+(-14)\cdot5^{k-1}\\
&=7\cdot12q+(7\cdot(-2))\cdot5^{k-1}\\
\end{align}
$7\cdot12q$ is a multiple of 7. Furthermore, $7\cdot(-2))\cdot5^{k-1}$ is a multiple of 7, thus their sum is a multiple of 7, illustrated by the right hand side of the equation.
The last line shows that $P(k+1)$ is true under the assumption that $P(k)$ is true. This completes the inductive step.
Thus by the Principal of Mathematical Induction, the initial proposition is true for all positive integers n. This completes the proof.
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