How do I show that $6(4^n-1)$ is a multiple of $9$ for all $n\in \mathbb{N}$? How do I show that $6(4^n-1)$ is a multiple of $9$ for all $n\in \mathbb{N}$? I'm not so keen on divisibility tricks. Any help is appreciated.

You want it to be a multiple of $9$, it suffices to show you can extract a pair of 3's from this. The $6$ has one of the 3's, and $4^n-1$ is 0 mod 3 so you're done.
