Question

provar que (2^n geq n + 1) para qualquer (n in mathbb{n}).

• (p(1)) é verdadeira, já que (2^1 geq 2).
• (p(n) rightarrow p(n+1)): assumindo, por hipótese de indução (hi), que para dado (n) (fixo) se tem (2^n geq n + 1) queremos provar que (2^{n+1} geq n + 2). então, da hi,

2^{n+1} = 2 cdot 2^n geq 2(n + 1) = 2n + 2 geq n + 2
já que (n > 0). por transitividade, (2^{n+1} geq n + 2), como queríamos mostrar.

Explanation:

🆕 New Concept Discovered: Mathematical Induction
Prove statements for all natural numbers in two steps.

Step 1: Understand the Goal

We want to prove the inequality:

$$ 2^n \geq n + 1 $$

for any natural number \( n \in \mathbb{N} \) (where \( \mathbb{N} = \{1, 2, 3, \dots\} \)).

To do this, we use **[LLM SSE On Failure]

Answer:

🆕 New Concept Discovered: Mathematical Induction
Prove statements for all natural numbers in two steps.

Step 1: Understand the Goal

We want to prove the inequality:

$$ 2^n \geq n + 1 $$

for any natural number \( n \in \mathbb{N} \) (where \( \mathbb{N} = \{1, 2, 3, \dots\} \)).

To do this, we use **[LLM SSE On Failure]