Proving a relation with induction I have a problem:
Let $p_n$ be the $n:th$ prime number ($p_1=2,  p_2=3,  p_3=5$ and so on). With induction, show that $p_{n+2}>3n$ for each integer $n\geq1$.
I can't figure this out because the primes are confusing me, making me unable to show the inductive step.

Proof by induction.
Let's start with the case $n=1$:
$p_{1+2}=p_3=5>3*1=3$
So the equality is true for $n=1$.
Now, we assume that the equality is true for the case $n-1$, that is,
$p_{n-1+2}=p_{n+1}>3(n-1)=3n-3$
and let's see that it is also true for n:
$p_{n+2}\geq p_{n+1}+2>3(n-1)+2=3n-1$; so $p_{n+2}\geq3n$
Now,you have to see that $p_{n+2}>3n$
