Let $f:\Bbb N\to\Bbb N$ be given by
$$f(n) = \begin{cases} 3, & n = 1 \\ 1, & n = 2 \\ 2f(n-1)+f(n-2), & n \ge 3 \end{cases}$$
prove that for $f(n)$ for all is odd all natural numbers $n$. (using complete induction)
I cant set this question up can someone help me. I know there are two cases even and odd numbers but I dont get complete induction
The Base case for n=1,2 is obvious.
Induction assumption: for all natural m < n, f(m) is odd.
Induction step:
f(n) = 2f(n-1) + f(n-2)
Since f(n-1) is natural, 2*f(n-1) is even (multiplied by two).
n-2 < n, and thus, from the induction assumption: f(n-2) is odd.
Adding an even number (2*f(n-1)) to an odd number (f(n-2)) gets an odd number, thus f(n) is odd.