If $P_n$ is the statement $3^n>n,$ prove $P_{n+1}$ is true whenever $P_n$ is true.

Question

If $3^n>n$ then $3^{n+1}>n+1$

I could not solve this problem

Answer 1

If $n=0$, then the truth of this statement is easily verifiable, since

$$3^0 = 1 > 0.$$

If $n>0$, i.e. $n\ge1$, then:

$$3^{n+1} = (by \: basic \: arithmetic)$$

$$3\cdot3^n > (by \: assumption \: 3^n > n)$$

$$3n = (by \: basic \: arithmetic)$$

$$2n + n \ge (by \: n \ge 1)$$

$$2n + 1 > (by \: n \ge 1)$$

$$n + 1$$

Answer 2

Presumably, $n\in\Bbb N$ since we're talking about an induction proof (you could also include $n=0$ but I won't here). Thus, $$n\geq 1\implies n+n\geq n+1\implies n+n+n>n+1\implies 3n>n+1$$ You also have $3^n>n$ by assumption. I'll leave the rest to you.