So, there is an excersise in my textbook:
Prove by induction
If $ P_n $ is n-th prime number and $ n\geqslant 12 $, than: $$P_n>3n$$
My approach:
1.For $n={12}:\quad P_{12}=37>3*12=36$
2.Inductive step:$\quad P_{n+1}>3n+3$
3.I figured out that:$\quad P_{n+1}-P_n\geqslant 2$
I got stuck, so i decided to go check answers: $$P_{n+1}\geqslant P_n+2>3n+2\geqslant 3n+3$$
That last part seems very wrong to me, and i do not think this is valid proof.I don't really know what to do next.
On
Here is an alternate approach using induction twice. Note all primes $\gt 3$ are congruent to either $1$ or $5$ modulo $6$. Thus, for all $n \gt 2$, the difference between $p_n$ and $p_{n+2}$ is at least $6 = 3 \times 2$. As such, you can use induction to prove the statement of
$$p_{n} \gt 3n \tag{1}\label{eq1A}$$
is true for all odd index primes (i.e, for $n = 2m+1$, you have $p_{2m+1} \gt 3(2m+1)$), and then also use it to prove it's true for all even index primes (i.e, for $n = 2m$, you have $p_{2m} \gt 3(2m)$), for all $n \ge 12$ in both cases. I'll leave it to you to do the rest yourself.
The way the chain of inequalities is written is wrong. Apparently, it should be$$P_{n+1}\ge P_{n}+2>3n+2\\\implies P_{n+1}\ge3n+3$$Now you should use the fact that $3\mid3n+3$ but $3\nmid P_{n+1}\ne3$. So $P_{n+1}\ne3n+3$.