Prove $9 \mathrel| (4^n+6n-1)$ by induction I know that this question was already answered, but I would like to know if the second step of induction its okay the way I did it.
This question is different from Induction proof for $n\in\mathbb N$, $9 \mathrel| (4^n+6n-1)$ and Let $n ∈ N, n ≥ 1$. Prove that $4^n + 6n - 1$ is divisible by $9$. since it is just about to check a part from a proof that I did (if the way I thought is correct or not).
Suppose that   $9 \mathrel| (4^k+6k-1)$ is true for $k\in \mathbb{N},  k\geq1$. I want to prove $P(k+1)$: $ 9 \mathrel| 4^{(k+1)} +6(k+1) - 1$
Since $9 \mathrel| (4^k+6k-1)$, I can rewrite as:
$4^k+6k-1 = 9q$,  where $q \in \mathbb{Z} $. By the definion of Divisibility.


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*Multiplying the equation on both sides by 4, I'll get:
$4^{(k+1)} +24k-4 = 36q$


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*Adding 3 on both sides
$4^{(k+1)} +24k - 1 = 36q + 3$


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*Subtracting $18k$ on both sides
$4^{(k+1)} +6k - 1 = 36q + 3 -18k$


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*Adding 6 in order to have $6(k+1)$ as a factor
$4^{(k+1)} +6k + 6 - 1 = 36q + 3 -18k + 6$
$4^{(k+1)} +6(k+1) - 1 = 36q + 9 -18k $
Note that $36q + 9 -18k$ can be written like this: $9(4q + 1 - 2k)$. Thus $9Q$, $Q=(4q + 1 - 2k), Q \in \mathbb{Z}$.
Therefore:


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*$4^{(k+1)} +6(k+1) - 1 = 9Q$. Which means that $4^{(k+1)} +6(k+1) - 1$ is also divided by 9.


Does my proof look fine? If there is a problem, please tell me. All help is appreciated.

If $n \equiv 0 \pmod 3$, then for an integer $k$:
$$4^n + 6n-1 \equiv 4^{3k} + 18k - 1 \equiv 1^k + 0 - 1 \equiv 0 \pmod 9.$$
If $n \equiv 1 \pmod 3$, then:
$$4^n + 6n-1 \equiv 4 \cdot4^{3k} + 18k + 6- 1 \equiv 4+0+6-1 \equiv 0 \pmod 9.$$
Finally, if $n \equiv 2 \pmod 3$, we have:
$$4^n + 6n-1 \equiv 16 \cdot4^{3k} + 18k + 12- 1 \equiv 16+0+12-1 \equiv 0 \pmod 9.$$
Since an integer can only congruent to either $0 \pmod 3, 1 \pmod 3$ or $2 \pmod 3$, this completes the proof for all integer $n$.
