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

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

My attempt:

Proof by Induction:

Base Case:

$ n = 1,  4^1 + 6(1) - 10 = 0 = 18(0)$. So base case holds.

I.H: Assume $ \ 4^{k} + 6k -10 = 18m$ for some $ m \in \mathbb{Z},\  k \in \mathbb{N}$.

WTP $ \ 4^{k+1} + 6(k+1) - 10 = 18m$.

$ 4^{k+1} + 6(k+1) - 10 = 4\cdot4^k +6k + 6 -10 = 4(18m + 10 -6k) + 6k + 6 -10$ , by I.H.

$ 4(18m + 10 -6k) + 6k + 6 -10 = 72m +40 - 18k -4 = 72m -18k +36 = 18(4m -k +2)$. 

Hence by induction, the result holds for all natural numbers. 

I have seen similar proofs in which instead of substituting an expression using the induction hypotheses, the $ k+1$ statement is manipulated until its brought in the desired form. Can I use the induction hypothesis to sub in $ \ 4^{k}$ like this? Is this approach correct?
Your proof looks okay.

Here is an alternate approach:

mod $18$, we have
$$
\begin{array}{c|c}
n&amp;1&amp;2&amp;3&amp;4&amp;\cdots\\
\hline
4^n&amp;4&amp;16&amp;10&amp;4&amp;\cdots\\
\hline
6n&amp;6&amp;12&amp;0&amp;6&amp;\cdots
\end{array}
$$
Since each column is dependent only on the previous one, the sequences for $4^n$ and $6n$ repeat every $3$ columns. Also from the table above,
$$
4^n+6n-10\equiv0\pmod{18}
$$
will also repeat every $3$ columns.
Nit-pick: Your WTP should be $4^{k+1}+6(k+1)-10=18\cdot m'$ for some $m' \in \mathbb{Z}$, rather than using the specific $m$ from the IH