Prove that if $n \in \mathbb{N}$ and $n \ge 2$, then $2^{n + 1} \le 3^n$.

Question

Prove that if $n \in \mathbb{N}$ and $n \ge 2$, then $2^{n + 1} \le 3^n$.

My method:

If $n = 2$, $2^{n + 1} \le 3^n$ then $2^3 \le 3^2$ is $8 \le 9$, which holds for $n = 2$.

$2^{k + 1} \le 3^k$ then $2^{k + 2} \le 3^{k + 1}$. Then $2 \cdot 2 \cdot 2^k \lt 3 \cdot 3^k$

$4 \cdot 2^k \lt 3 \cdot 3^k$

Therefore, $4 \cdot 2^k \lt 3 \cdot 3^k$. Hence, $2^{n + 1} \le 3^n$ for $n \ge 2$.

Is there a problem with: $2^{k + 1} \le 3^k$ then $2^{k + 2} \le 3^{k + 1}$ or any other part of this proof?

Answer 1

The proof is easy:

• If $k=2$, we know that $2^{k+1} = 2^3 = 8 \leq 9 = 3^2 = 3^k$. Then, it is true.
• We assume that it is also true for $k=n$, then we assume that $2^{n+1} \leq 3^{n}$
• We want to see that it is true if $k = n+1$:

$2^{k+1} = 2^{n+2} = 2^{n+1} \cdot 2$

We assumed before $2^{n+1} \leq 3^{n}$, then

$2^{k+1} = 2^{n+1} \cdot 2 \leq 3^{n} \cdot 2 \leq 3^{n} \cdot 3 = 3^{n+1}=3^k \rightarrow 2^{k+1} \leq 3^k$

What we do is:

• We prove that it is true for $k=2$
• We prove that if it is true for $k=n$, then it is also true for $k=n+1$

Therefore, it is true for $k=3$ because it is true for $k=2$. It is true for $k=4$ because it is true for $k=3$...

Answer 2

Formal proof by induction.

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First, show that this is true for $n=2$:

$2^{2+1}<3^2$

Second, assume that this is true for $n$:

$2^{n+1}<3^n$

Third, prove that this is true for $n+1$:

$2^{n+2}=$

$2\cdot\color\red{2^{n+1}}<$

$2\cdot\color\red{3^n}<$

$3\cdot3^n=$

$3^{n+1}$

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Please note that the assumption is used only in the part marked red.