Big-O of recursive function Let $f:\mathbb{Z}_+ \to \mathbb{Z}_+$ be the function defined by $f(k)=3f(k-1)+2$ for any $k \in \mathbb{Z}_+$. Prove that $f(n)$ is $O(6^n)$.
How do I prove it with mathematical induction?

from iteration :
$f(n)=3f(n-1)+2$
...
$\space\space\space\space\space\space\space\space=3^if(n-i-1)+2+6+18+...+2*3^i$
(n-i-1)=1 => i=n-2
$f(n)=3^{n-2}f(1)+\sum _{n=0}^{i}2*3^n$
$\sum_{n=0}^{i}2*3^n= {2(1-2*3^i) \over -2}=-(1-2*3^i) $
$f(n)=3^{n-2}-(1-2*3^{n-2})=-1+3^{n-2}(3)=-1+3^{n-1}$
induction :
n=0
$-1-3^{-1} $is O(1)
n=k
$-1+3^{k-1}=O(6^{k})$ => $3^{k-1}=O(6^{k})=> 6*3^k\le 3c6^{k+1}$ 
n=k+1
$-1+3^{k}\le3^{k}\le6*3^k\le 3c 6^{k+1}=> -1+3^k\le c'6^{k+1}=> -1+3^k=O(6^{k+1})$
