Choose the proposition among the following that is not true for `n=1` but true for all `n gt 1`.

Options

A. `n! gt ((n+1)/(2))^(n)`
B. `n! gt ((n+1)/(2))^(n)` is an integer
C. `(d^(n))/(dx^(n))(sin ax+b)=a^(n)sin(ax+b+(npi)/(2))`
D. `n! lt ((n)/(2))^(n)`

Answer

To solve the problem, we need to analyze the propositions given and determine which one is not true for \( n = 1 \) but true for all \( n > 1 \). Let's go through the propositions step by step.

### Step 1: Analyze Proposition 1
**Proposition 1:** \( n! > \frac{n + 1}{2^n} \)

- For \( n = 1 \):
- LHS: \( 1! = 1 \)
- RHS: \( \frac{1 + 1}{2^1} = \frac{2}{2} = 1 \)

Here, \( 1 \) is not greater than \( 1 \), so this proposition is **not true for \( n = 1 \)**.

- For \( n > 1 \):
- We can use mathematical induction to prove that \( n! > \frac{n + 1}{2^n} \) holds true for \( n > 1 \).

### Step 2: Induction Step for Proposition 1
1. **Base Case:** For \( n = 2 \):
- LHS: \( 2! = 2 \)
- RHS: \( \frac{2 + 1}{2^2} = \frac{3}{4} \)
- \( 2 > \frac{3}{4} \) is true.

2. **Inductive Hypothesis:** Assume \( k! > \frac{k + 1}{2^k} \) is true for some \( k > 1 \).

3. **Inductive Step:** Show it holds for \( k + 1 \):
- LHS: \( (k + 1)! = (k + 1) \cdot k! \)
- By the inductive hypothesis: \( k! > \frac{k + 1}{2^k} \)
- Thus, \( (k + 1)! > (k + 1) \cdot \frac{k + 1}{2^k} = \frac{(k + 1)^2}{2^k} \)

We need to show that:
\[ \frac{(k + 1)^2}{2^k} > \frac{(k + 2)}{2^{k + 1}} \]
Simplifying gives:
\[ 2(k + 1)^2 > k + 2 \]
This inequality holds for \( k > 1 \).

Thus, Proposition 1 is true for all \( n > 1 \).

### Step 3: Analyze Proposition 2
**Proposition 2:** \( n! > \frac{n + 1}{2^n} \) (with the condition that \( n \) is an integer)

This is essentially the same as Proposition 1. Since we already established that it is not true for \( n = 1 \) and true for \( n > 1 \), we can conclude the same.

### Step 4: Analyze Proposition 3
**Proposition 3:** The nth derivative of \( \sin(ax + b) \) is given by \( a^n \sin(ax + b + n\frac{\pi}{2}) \).

- For \( n = 1 \):
- LHS: \( \frac{d}{dx} \sin(ax + b) = a \cos(ax + b) \)
- RHS: \( a \sin(ax + b + \frac{\pi}{2}) = a \cos(ax + b) \)

This holds true for \( n = 1 \).

- For \( n > 1 \):
- The pattern continues, and the formula holds for all integers \( n \).

### Step 5: Analyze Proposition 4
**Proposition 4:** \( n! < \frac{n}{2^n} \)

- For \( n = 1 \):
- LHS: \( 1! = 1 \)
- RHS: \( \frac{1}{2^1} = \frac{1}{2} \)

Here, \( 1 < \frac{1}{2} \) is **not true**.

- For \( n > 1 \):
- Checking for \( n = 2 \):
- LHS: \( 2! = 2 \)
- RHS: \( \frac{2}{2^2} = \frac{2}{4} = \frac{1}{2} \)

\( 2 < \frac{1}{2} \) is also **not true**.

### Conclusion
The propositions that are not true for \( n = 1 \) but true for \( n > 1 \ are Proposition 1 and Proposition 2. However, both propositions are essentially the same.