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

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.

let $$T(n)=4^n+6n-10$$ then $$T(n+1)=4^{n+1}+6(n+1)-10$$ and we get $$T(n+1)-T(n)=3(4^n+2)$$ thus $$T(n+1)=T(n)+3(4^n+2)$$ it is clear then $3|4^n+2$ so $9|3(4^n+2)$ and $4^n+2$ is even (note that $$18|T(n)$$ via induction