`""^(m)C_(r+1)+ sum_(k=m)^(n)""^(k)C_(r)` is equal to :

Options

A. `""^(n)C_(r+1)`
B. `""^(n +1)C_(r +1)`
C. `""^(n)C_(r-1)`
D. `""^(n)C_(r-1)`

Answer

To solve the problem \( \binom{m}{r+1} + \sum_{k=m}^{n} \binom{k}{r} \), we will use the Hockey Stick Identity (or Christmas Stocking Theorem) in combinatorics.

### Step-by-step Solution:

1. **Identify the components of the expression**:
We have two parts in the expression:
- The first part is \( \binom{m}{r+1} \).
- The second part is the summation \( \sum_{k=m}^{n} \binom{k}{r} \).

2. **Apply the Hockey Stick Identity**:
The Hockey Stick Identity states that:
\[ \sum_{j=r}^{n} \binom{j}{r} = \binom{n+1}{r+1} \]
In our case, we need to adjust the limits of the summation. We can rewrite the summation as:
\[ \sum_{k=m}^{n} \binom{k}{r} = \sum_{j=m}^{n} \binom{j}{r} \]

3. **Separate the summation**:
We can express the summation from \( m \) to \( n \) as:
\[ \sum_{k=m}^{n} \binom{k}{r} = \binom{m}{r} + \sum_{k=m+1}^{n} \binom{k}{r} \]

4. **Use the Hockey Stick Identity on the summation**:
According to the Hockey Stick Identity, we can express the summation as:
\[ \sum_{k=m}^{n} \binom{k}{r} = \binom{n+1}{r+1} \]

5. **Combine the results**:
Now we can combine the results:
\[ \binom{m}{r+1} + \sum_{k=m}^{n} \binom{k}{r} = \binom{m}{r+1} + \binom{n+1}{r+1} \]

6. **Final result**:
The final expression can be simplified further using the Hockey Stick Identity:
\[ \binom{m}{r+1} + \binom{n+1}{r+1} = \binom{n+1}{r+1} + \binom{m}{r+1} = \binom{n+m+1}{r+1} \]

Thus, the final answer is:
\[ \binom{n+1}{r+1} \]