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.

Try demonstration by induction. Show the case when $n = 1$, then assume that for $n = k$ the theorem is true and then try demonstrate that the result of $n = k + 1$ is multiple of 9.
