A sequence divisible by 9 I was trying to solve this series by mathematical induction for every $n$ from $\Bbb N$  : $u_n=n4^{n+1}-(n+1)4^n+1$ is divisible by $9$. 
The initiation was pretty easy, but I only managed to prove $u_{n+1}=3k$ while $k$ is an integer and I don't think if it's divisible by $3$ implies that it is divisible by $9$ ; is it ? if not how can I proceed to prove the divisibility ? by mod maybe? thanks in advance for your answer

For integer $n\ge0,$ using Binomial Expansion
$$4^m=(1+3)^m\equiv1+3m\pmod9$$
$$\implies u_n=n4^{n+1}-(n+1)4^n+1$$
$$\equiv n\{1+3(n+1)\}-(n+1)(1+3n)+1\pmod9$$
$$\equiv4n+3n^2-(4n+1+3n^2)+1\equiv0$$
