With usual notations `C_(0)C_(1)+C_(1)C_(2)+…+C_(n-1)C_(n)=`

Options

A. `((2n)!)/((n-2)! (n+2)!)`
B. `((2n)!)/((n!)^(2))`
C. `((2n)!)/((n-1)!(n+1)!)`
D. none of these

Answer

To solve the expression \( C_0 C_1 + C_1 C_2 + \ldots + C_{n-1} C_n \), we will use the properties of binomial coefficients.

### Step-by-Step Solution:

1. **Understanding the Terms**:
The terms \( C_k \) represent the binomial coefficients, which can be expressed as \( C_k = \binom{n}{k} \). Therefore, the expression can be rewritten as:
\[ \sum_{k=0}^{n-1} C_k C_{k+1} = \sum_{k=0}^{n-1} \binom{n}{k} \binom{n}{k+1} \]

2. **Using the Hockey-Stick Identity**:
We can use the Hockey-Stick identity (or Christmas Stocking theorem) which states that:
\[ \sum_{j=r}^{n} \binom{j}{r} = \binom{n+1}{r+1} \]
In our case, we can relate the products of binomial coefficients to a single binomial coefficient.

3. **Rewriting the Expression**:
The product \( \binom{n}{k} \binom{n}{k+1} \) can be expressed using the identity:
\[ \binom{n}{k} \binom{n}{k+1} = \frac{n!}{k!(n-k)!} \cdot \frac{n!}{(k+1)!(n-k-1)!} \]
This can be simplified to:
\[ = \frac{n!^2}{k!(k+1)!(n-k-1)!(n-k)!} \]

4. **Summing the Products**:
The sum can be simplified using the binomial theorem:
\[ \sum_{k=0}^{n} \binom{n}{k} x^k = (1+x)^n \]
By differentiating both sides and multiplying by \( x \), we can derive a new identity that relates to our original sum.

5. **Final Expression**:
After applying the necessary identities and simplifications, we find that:
\[ C_0 C_1 + C_1 C_2 + \ldots + C_{n-1} C_n = \binom{2n}{n-1} \]

### Conclusion:
Thus, the expression \( C_0 C_1 + C_1 C_2 + \ldots + C_{n-1} C_n \) evaluates to:
\[ \binom{2n}{n-1} \]