If `C_(r )= ((10),(r))`, then `Sigma_(r=1)^(10)C_(r-1) C_(r)` is equal to

Options

A. `((20),(9))`
B. `((20),(10))`
C. `((20),(13))`
D. `((20),(8))`

Answer

To solve the problem, we need to evaluate the expression \( \Sigma_{r=1}^{10} C_{r-1} C_r \) where \( C_r = \binom{10}{r} \).

### Step-by-Step Solution:

1. **Understanding the Binomial Coefficients**:
We have \( C_r = \binom{10}{r} \). Therefore, we can rewrite the expression:
\[ \Sigma_{r=1}^{10} C_{r-1} C_r = \Sigma_{r=1}^{10} \binom{10}{r-1} \binom{10}{r} \]

2. **Changing the Index of Summation**:
To simplify the summation, we can change the index. Let \( k = r - 1 \). Then, when \( r = 1 \), \( k = 0 \) and when \( r = 10 \), \( k = 9 \). Thus, the summation becomes:
\[ \Sigma_{k=0}^{9} \binom{10}{k} \binom{10}{k+1} \]

3. **Using the Hockey-Stick Identity**:
We can use the identity:
\[ \sum_{k=r}^{n} \binom{n}{k} \binom{k}{r} = \binom{n+1}{r+1} \]
In our case, we will apply it as follows:
\[ \sum_{k=0}^{9} \binom{10}{k} \binom{10}{k+1} = \binom{11}{11} \]

4. **Simplifying the Result**:
The expression simplifies to:
\[ \binom{11}{11} = 1 \]

Thus, the final result of the summation \( \Sigma_{r=1}^{10} C_{r-1} C_r \) is:
\[ \boxed{1} \]