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.
 A: OP says that this is accepted as an answer, so I post it, for the sake of convenience.
Hint
$$4^n=(1+3)^n=1+\Sigma_{k=1}^n\binom{n}{k}3^k,$$ so $4^n-1=\Sigma_{k=1}^n\binom{n}{k}3^k$ is divisible by $3$.
Barring mistakes, and thanks.
A: 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.
A: 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.
A: The following identity is often useful:
$$x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1).$$
Like many identities, it is easy to verify: just expand the right-hand side, and note that almost everything cancels. 
If we put $x=4$, then it follows from the above identity that $4-1$ divides $4^n-1$.
Remark: For $x\ne 1$, the identity can be rewritten as 
$$1+x+x^2+\cdots +x^{n-1}=\frac{x^n -1}{x-1}.$$ 
This is one version of the formula for the sum of a finite geometric series. Everything is connected to everything else.
