The value of the natural numbers n such that `2 ^(n) gt 2n +1 ` is valid for

Options

A. `n ge 3`
B. `n lt 3`
C. `n gt 1`
D. any n

Answer

To solve the inequality \( 2^n > 2n + 1 \) for natural numbers \( n \), we can follow these steps:

### Step 1: Analyze the inequality
We need to find the values of \( n \) such that \( 2^n > 2n + 1 \).

### Step 2: Test small values of \( n \)
Let's start by testing some small natural numbers:

- For \( n = 1 \):
\[ 2^1 = 2 \quad \text{and} \quad 2 \cdot 1 + 1 = 3 \]
So, \( 2 > 3 \) is **not valid**.

- For \( n = 2 \):
\[ 2^2 = 4 \quad \text{and} \quad 2 \cdot 2 + 1 = 5 \]
So, \( 4 > 5 \) is **not valid**.

- For \( n = 3 \):
\[ 2^3 = 8 \quad \text{and} \quad 2 \cdot 3 + 1 = 7 \]
So, \( 8 > 7 \) is **valid**.

- For \( n = 4 \):
\[ 2^4 = 16 \quad \text{and} \quad 2 \cdot 4 + 1 = 9 \]
So, \( 16 > 9 \) is **valid**.

### Step 3: Generalize the results
From our tests, we see that the inequality holds for \( n = 3 \) and \( n = 4 \). Let's check if it holds for larger values of \( n \):

- For \( n = 5 \):
\[ 2^5 = 32 \quad \text{and} \quad 2 \cdot 5 + 1 = 11 \]
So, \( 32 > 11 \) is **valid**.

- For \( n = 6 \):
\[ 2^6 = 64 \quad \text{and} \quad 2 \cdot 6 + 1 = 13 \]
So, \( 64 > 13 \) is **valid**.

### Step 4: Conclusion
As we can see, the inequality \( 2^n > 2n + 1 \) holds for \( n = 3 \) and all natural numbers greater than 3. Therefore, the final conclusion is:

The values of natural numbers \( n \) such that \( 2^n > 2n + 1 \) is valid for \( n \geq 3 \).