If n be a positive integer and `P_n` denotes the product of the binomial coefficients in the expansion of `(1 +x)^n`, prove that `(P_(n+1))/P_n=(n+1)^n/(n!)`.

Options

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

Explanation

To prove that \(\frac{P_{n+1}}{P_n} = \frac{(n+1)^n}{n!}\), where \(P_n\) denotes the product of the binomial coefficients in the expansion of \((1+x)^n\), we will follow these steps:

### Step 1: Define \(P_n\) and \(P_{n+1}\)

The product of the binomial coefficients in the expansion of \((1+x)^n\) is given by:
\[ P_n = \prod_{k=0}^{n} \binom{n}{k} \]
Similarly, for \(P_{n+1}\):
\[ P_{n+1} = \prod_{k=0}^{n+1} \binom{n+1}{k} \]

### Step 2: Express \(P_{n+1}\) in terms of \(P_n\)

We can express \(P_{n+1}\) as:
\[ P_{n+1} = \prod_{k=0}^{n} \binom{n+1}{k} \cdot \binom{n+1}{n+1} \]
Since \(\binom{n+1}{n+1} = 1\), we have:
\[ P_{n+1} = \prod_{k=0}^{n} \binom{n+1}{k} \]

### Step 3: Use the property of binomial coefficients

Using the identity for binomial coefficients:
\[ \binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1} \]
We can express \(\binom{n+1}{k}\) in terms of \(\binom{n}{k}\) and \(\binom{n}{k-1}\).

### Step 4: Calculate the ratio \(\frac{P_{n+1}}{P_n}\)

Now, we can write:
\[ \frac{P_{n+1}}{P_n} = \frac{\prod_{k=0}^{n} \binom{n+1}{k}}{\prod_{k=0}^{n} \binom{n}{k}} \]
This simplifies to:
\[ \frac{P_{n+1}}{P_n} = \prod_{k=0}^{n} \frac{\binom{n+1}{k}}{\binom{n}{k}} \]

### Step 5: Simplify the ratio of binomial coefficients

Using the definition of binomial coefficients:
\[ \frac{\binom{n+1}{k}}{\binom{n}{k}} = \frac{(n+1)!/(k!(n+1-k)!)}{n!/(k!(n-k)!)} = \frac{(n+1)(n-k)!}{(n+1-k)!} \]
This can be further simplified to:
\[ \frac{(n+1)}{(n+1-k)} \]

### Step 6: Calculate the product

Thus, we have:
\[ \frac{P_{n+1}}{P_n} = \prod_{k=0}^{n} \frac{(n+1)}{(n+1-k)} = (n+1)^n \cdot \frac{1}{n!} \]

### Step 7: Final result

Therefore, we conclude that:
\[ \frac{P_{n+1}}{P_n} = \frac{(n+1)^n}{n!} \]
This completes the proof.