If `C_(r) = ""^(n)C_(r) and (C_(0) + C_(1)) (C_(1) + C_(2)) … (C_(n-1) + C_(n)) = `
`k ((n +1)^(n))/(n!)`, then the value of k, is

Options

A. `C_(0)C_(1) C_(2)…C_(n)`
B. `C_(1)^(2) C_(2)^(2) ….C_(n)^(2)`
C. `C_(1) + C_(2) +…+ C_(n)`
D. none of these

Answer

To solve the problem, we need to evaluate the expression \((C_0 + C_1)(C_1 + C_2) \cdots (C_{n-1} + C_n)\) and find the value of \(k\) such that:

\[ (C_0 + C_1)(C_1 + C_2) \cdots (C_{n-1} + C_n) = k \cdot \frac{(n+1)^n}{n!} \]

### Step 1: Understand the Binomial Coefficient Identity

We know from the properties of binomial coefficients that:

\[ C_r + C_{r-1} = C_{r+1} \]

This means we can rewrite each term in the product.

### Step 2: Rewrite Each Term

Using the identity, we can rewrite the terms in the product:

\[ C_0 + C_1 = C_1 + C_0 = C_1 \]
\[ C_1 + C_2 = C_2 + C_1 = C_2 \]
\[ C_2 + C_3 = C_3 + C_2 = C_3 \]
\[ \vdots \]
\[ C_{n-1} + C_n = C_n + C_{n-1} = C_n \]

Thus, we can express the entire product as:

\[ (C_0 + C_1)(C_1 + C_2) \cdots (C_{n-1} + C_n) = C_1 \cdot C_2 \cdots C_n \]

### Step 3: Express the Product in Terms of \(n+1\)

We can express the product \(C_0 + C_1\) through \(C_n\) using the identity:

\[ C_0 + C_1 = C_1, \quad C_1 + C_2 = C_2, \quad \ldots, \quad C_{n-1} + C_n = C_n \]

This gives us:

\[ (C_0 + C_1)(C_1 + C_2) \cdots (C_{n-1} + C_n) = C_1 \cdot C_2 \cdots C_n \]

### Step 4: Calculate the Product

The product of the coefficients can be expressed as:

\[ C_0 C_1 C_2 \cdots C_n = \frac{(n+1)!}{1! \cdot 2! \cdots n!} \]

### Step 5: Relate to the Given Expression

We need to relate this to the expression given in the problem:

\[ C_0 C_1 C_2 \cdots C_n = \frac{(n+1)^n}{n!} \]

### Step 6: Find \(k\)

From the equation:

\[ (C_0 + C_1)(C_1 + C_2) \cdots (C_{n-1} + C_n) = k \cdot \frac{(n+1)^n}{n!} \]

We can see that:

\[ k = C_0 C_1 C_2 \cdots C_n \]

### Final Value of \(k\)

Since \(C_n = 1\), we find that:

\[ k = 1 \]

Thus, the value of \(k\) is:

\[ \boxed{1} \]