`sum_(r=0)^m "^(n+r) C_n` is equal to

Options

A. `"^(n+m+1) C_(n-1)`
B. `"^(n+2+2) C_n`
C. `"^(n+m+3) C_(n-1)`
D. none of these

Answer

To solve the problem \( \sum_{r=0}^{m} \binom{n+r}{n} \), we will follow these steps:

### Step 1: Understanding the Summation
We need to evaluate the summation from \( r = 0 \) to \( m \) of the binomial coefficient \( \binom{n+r}{n} \). This coefficient represents the number of ways to choose \( n \) items from \( n+r \) items.

### Step 2: Expanding the Summation
The summation can be written explicitly as:
\[ \sum_{r=0}^{m} \binom{n+r}{n} = \binom{n+0}{n} + \binom{n+1}{n} + \binom{n+2}{n} + \ldots + \binom{n+m}{n} \]

### Step 3: Using the Hockey-Stick Identity
We can apply the Hockey-Stick Identity (or Christmas Stocking Theorem) from combinatorics, which states that:
\[ \sum_{i=k}^{r} \binom{i}{k} = \binom{r+1}{k+1} \]
In our case, we can set \( k = n \) and \( r = n + m \). Therefore, we can rewrite our summation as:
\[ \sum_{r=0}^{m} \binom{n+r}{n} = \sum_{i=n}^{n+m} \binom{i}{n} = \binom{n+m+1}{n+1} \]

### Step 4: Final Result
Thus, the final result of the summation is:
\[ \sum_{r=0}^{m} \binom{n+r}{n} = \binom{n+m+1}{n+1} \]

### Step 5: Verifying the Result
To verify, we can observe that the result can also be expressed in terms of \( n+m+1 \) choose \( n \) minus 1:
\[ \binom{n+m+1}{n} = \binom{n+m+1}{n+1} \]
This confirms that our result is consistent with the Hockey-Stick Identity.

### Conclusion
The final answer to the summation \( \sum_{r=0}^{m} \binom{n+r}{n} \) is:
\[ \binom{n+m+1}{n+1} \]