If `P_(n)` denotes the product of the binomial coefficients in the expansion of `(1+x)^(n)`, then `(P_(n+1))/(P_(n))` equals

Options

A. `((n+1)^(n))/(n !)`
B. `(n^(n))/(n !)`
C. `((n+1)^(n))/((n+1)!)`
D. `((n+1)^(n+1))/((n+1)!)`

Answer

To solve the problem, we need to find the ratio \( \frac{P_{n+1}}{P_n} \), where \( P_n \) is the product of the binomial coefficients in the expansion of \( (1+x)^n \).

### Step-by-Step Solution:

1. **Understanding the Binomial Coefficients**:
The binomial coefficients in the expansion of \( (1+x)^n \) are given by:
\[ \binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n} \]
Therefore, the product \( P_n \) can be expressed as:
\[ P_n = \binom{n}{0} \cdot \binom{n}{1} \cdot \binom{n}{2} \cdots \binom{n}{n} \]

2. **Expressing \( P_{n+1} \)**:
Similarly, for \( P_{n+1} \), we have:
\[ P_{n+1} = \binom{n+1}{0} \cdot \binom{n+1}{1} \cdot \binom{n+1}{2} \cdots \binom{n+1}{n+1} \]

3. **Writing the Ratio**:
We need to find:
\[ \frac{P_{n+1}}{P_n} = \frac{\binom{n+1}{0} \cdot \binom{n+1}{1} \cdots \binom{n+1}{n+1}}{\binom{n}{0} \cdot \binom{n}{1} \cdots \binom{n}{n}} \]

4. **Using the Binomial Coefficient Identity**:
We can use the identity:
\[ \binom{n+1}{r} = \binom{n}{r} + \binom{n}{r-1} \]
This means that each binomial coefficient \( \binom{n+1}{r} \) can be expressed in terms of \( \binom{n}{r} \) and \( \binom{n}{r-1} \).

5. **Calculating the Ratio**:
We can express \( P_{n+1} \) in terms of \( P_n \):
\[ P_{n+1} = \binom{n+1}{0} \cdot \binom{n+1}{1} \cdots \binom{n+1}{n} \cdot \binom{n+1}{n+1} \]
This can be rearranged to show that:
\[ \frac{P_{n+1}}{P_n} = \frac{\binom{n+1}{0} \cdot \binom{n+1}{1} \cdots \binom{n+1}{n}}{P_n} \cdot \binom{n+1}{n+1} \]

6. **Final Expression**:
After simplifying, we find that:
\[ \frac{P_{n+1}}{P_n} = \frac{(n+1)^{n+1}}{(n+1) \cdot n!} = \frac{(n+1)^n}{n!} \]

### Final Answer:
\[ \frac{P_{n+1}}{P_n} = \frac{(n+1)^n}{n!} \]