Prove that $\forall n \in \mathbb{N}: 4^n + 6n - 10$ is divisible by $18$ 
Prove that $\forall n \in \mathbb{N}: 4^n + 6n - 10$ is divisible by $18$

Base case: for $n = 1: 4^1 +6\cdot 1 - 10 = 0$ is divisible by 18.
Inductive Assumption: Assume that for for some $k \in \mathbb{N} :4^k +6k-10$
Proving that $4^{k+1}+6(k+1)-10$ is divisible by 18
$4^{k+1}+6(k+1)-10= 4^k \cdot 4 + 6k + 6 - 10$ $$= \color{green}{4^k +6k-10} + 3
 \cdot 4^k + 6$$ 
The first term is divisible by 18 according to the inductive assumption and I have to find a way to manipulate the second term to be divisible by 18.
There is a similar question posted, but the answers for it are substituting another expression $18m$ instead of manipulating $4^{k+1}+6(k+1)-10$. I want to know how to solve this without substituting 

\begin{align}3\cdot 4^k+6 &= 6(2\cdot4^{k-1}+1) \\
&=6(3\cdot4^{k-1}+1-4^{k-1})\\
&=6(3\cdot4^{k-1}+(1-2^{k-1})(1+2^{k-1}))\\
&=6(3\cdot4^{k-1}-(2^{k-1}-1)(1+2^{k-1}))\end{align}
Notice that $2^{k-1}$ is not divisible by $3$, hence either $(2^{k-1}-1)$ is divisible by $3$ or $(2^{k-1}+1)$ is divisible by $3$. 
Hence $(3\cdot4^{k-1}-(2^{k-1}-1)(1+2^{k-1}))$ is divisble by $3$ and $6(3\cdot4^{k-1}-(2^{k-1}-1)(1+2^{k-1}))$ is divisible by $18$.
Remark: Using modulo arithmetic makes the task much easier.
