Function and Arithmetic progressions.

Question

$f: \mathbb{N} -› \mathbb{N}$

$f(1)=2$ and $f(n+1)=f(n)+4$ For $n \ge 1$. Calculate $f(135)$

I don't know how to start to solve

Answer 1

$f(n+1)=f(n)+4$
$ \Rightarrow f(n+1) -f(n ) = 4 $ this means $f(n) $ is arithmetic progression with common difference 4 and whose first term is 2. Thus $f(135 ) = 2 + (135-1)4 = 538$