If `P_(n)` denotes the product of first n natural numbers, then for all `n inN.`

Options

A. `P_(n)len^(n)`
B. `P_(n)+1len^(n)`
C. `P_(n)le((n+1)/(2))^(n)`
D. none of these

Explanation

To solve the problem, we need to establish a relationship between \( P_n \) (the product of the first \( n \) natural numbers) and \( n \). We will use the Arithmetic Mean-Geometric Mean (AM-GM) inequality for this purpose.

### Step-by-Step Solution:

1. **Define \( P_n \)**:
\[ P_n = 1 \times 2 \times 3 \times \ldots \times n \]
This is the product of the first \( n \) natural numbers.

2. **Apply the AM-GM Inequality**:
According to the AM-GM inequality, for any set of positive numbers, the arithmetic mean is greater than or equal to the geometric mean. Therefore, we can write:
\[ \frac{1 + 2 + 3 + \ldots + n}{n} \geq \sqrt[n]{P_n} \]

3. **Calculate the Arithmetic Mean**:
The sum of the first \( n \) natural numbers is given by:
\[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]
Thus, the arithmetic mean becomes:
\[ \frac{1 + 2 + 3 + \ldots + n}{n} = \frac{\frac{n(n + 1)}{2}}{n} = \frac{n + 1}{2} \]

4. **Substitute into the AM-GM Inequality**:
Now substituting the arithmetic mean back into the inequality, we have:
\[ \frac{n + 1}{2} \geq \sqrt[n]{P_n} \]

5. **Raise Both Sides to the Power of \( n \)**:
To eliminate the root, we raise both sides to the power of \( n \):
\[ \left(\frac{n + 1}{2}\right)^n \geq P_n \]

6. **Rearranging the Inequality**:
This can be rearranged to show:
\[ P_n \leq \left(\frac{n + 1}{2}\right)^n \]

### Conclusion:
Thus, we have established that for all \( n \in \mathbb{N} \):
\[ P_n \leq \left(\frac{n + 1}{2}\right)^n \]