<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] For $n=1$ we have $4^1+6(1)-10=0$, which is divisible by $18$. If there exists $k \in \mathbb{N}$ such that $4^n+6n-10=18k$, for some $n \in \mathbb{N}$, then $$4^{n+1}+6(n+1)-10=4\cdot 4^n+6n+6-10=(4\cdot 4^n+24n-40)-24n+40+6n-4=4\cdot 18k-18n+36=18(k-n+2),$$ and since $m-n+2 \in \mathbb{Z}$ you have that $4^{n+1}+6(n+1)-10$ is divisible by 18. 
By induction, the result follows. 
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