Prove: For any integer $n \geq 2$, there is an odd number P such that $2n \lt P \lt 3n$ I am in high school and had this for a homework problem.  I got it wrong, but the teacher did not post the correct answer.  Any help would be appreciated.  It is about writing proofs.
Prove that for any integer n greater than or equal to 2, there is always an odd number P between 2n and 3n.

Write numbers between $2n$ and $3n$
$$
2n<2n+1 ,2n+1 ,2n+3 ,...,2n+(n-1)<3n 
$$
if $n>2$ 
there is $n-1$ numbers between $2n$, $3n$ 
number of terms=
$$ \frac{\text{last} -\text{first}}{\text{step}}+1=\\\frac{3n -2n}{1}+1=n+1\\$$
between $2n$, $3n$ are $(n+1)-2$ terms.
