Prove by induction that $2^{n} < 3^{n}, \forall n \in \mathbb{N}$

Question

PROBLEM:
Prove by induction that: $$ 2^{n} < 3^{n} , \forall n \in \mathbb{N} $$

ATTEMPT TO PROVE:

1. Let n=1. So, $2^1 = 2$ and $3^1 = 3$ (Clearly, $2<3$ )
2. Assume true at $n=k$. So, $2^k < 3^k$
3. Now, let $n=k+1$. (And this is where I get stuck) So,
   $$ 2^{k+1} = 2*2^k < 2*3^k = ? $$ I know that I need to manipulate $2*3^k$ so that I end up with $3^{k+1}$
   I was thinking I could do the following, but I don't know if it makes sense:
   $$ 2^{k+1} = 2*2^k < 2*3^k $$ $$ 2*3^k = 6^k $$ $$ 6^k = (3+3)^k $$ $$ (3+3)^k = 3^k + 3^k $$ $$ 3^k + 3^k = 3^{k+1} $$

Answer 1

What you have written is not correct. The way to proceed might be so simple that it's completely eluding you. This happens to all of us. You know that $2 < 3$, so $2\cdot 3^k < 3\cdot 3^k$.

Answer 2

Some general notes – you should always start with a proposition because it becomes unclear with harder induction problems what exactly you're trying to prove. Say $P(n) = 2^n < 3^n$.

Your base case is correct, so I will expand on the inductive step. You need to prove $P(n + 1)$ or $2^{n +1} < 3^{n + 1}$.

1. Assume $P(n)$ or $2^n < 3^n$
2. $2 \cdot 2^n < 2 \cdot 3^n$ (by the inductive hypothesis)
3. $2 \cdot 3^n < 3 \cdot 3^n$, so $2 \cdot 2^n < 3 \cdot 3^n$
4. $2 \cdot 2^n = 2^{n + 1}$ and $3 \cdot 3^{n} = 3^{n + 1}$, so $2^{n + 1} = 3^{n + 1}$