<s>[INST] 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 
 [/INST] HINT
If $4^k + 6k -10$ is divisible by $18$, then $4(4^k + 6k -10)$ is divisible by $18$ as well.
Subtract the resulting expression from the expression you get when plugging in $k+1$:
$$4^{k+1} +6(k+1)-10 - 4(4^k + 6k -10) = $$
$$4 \cdot 4^k +6k+6-10 - 4 \cdot 4^k -24k + 40 = $$
$$-18k+36$$
Since $-18k+36$ is clearly divisible by $18$, and since $4(4^k + 6k -10)$ is divisible by $18$, it follows that $4^{k+1} + 6(k+1) -10$ is divisible by $18$ as well.
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