The value of `sum_(i=0)^(n)""^(n-i)C_(r),rin[1,n]capsquare` is equal to :

Options

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

Answer

To solve the problem, we need to evaluate the summation \( \sum_{i=0}^{n} \binom{n-i}{r} \) where \( r \) is in the range [1, n].

### Step-by-step Solution:

1. **Understanding the Summation**:
The summation can be rewritten by changing the index of summation. Let \( j = n - i \). Then when \( i = 0 \), \( j = n \) and when \( i = n \), \( j = 0 \). Therefore, we can rewrite the summation as:
\[ \sum_{i=0}^{n} \binom{n-i}{r} = \sum_{j=0}^{n} \binom{j}{r} \]

2. **Using the Hockey Stick Identity**:
The Hockey Stick Identity in combinatorics states that:
\[ \sum_{k=r}^{n} \binom{k}{r} = \binom{n+1}{r+1} \]
In our case, we need to adjust the limits of the summation. Notice that the summation \( \sum_{j=0}^{n} \binom{j}{r} \) can be expressed as:
\[ \sum_{j=r}^{n} \binom{j}{r} + \sum_{j=0}^{r-1} \binom{j}{r} \]
The second term \( \sum_{j=0}^{r-1} \binom{j}{r} \) is zero because \( \binom{j}{r} = 0 \) for \( j < r \).

3. **Applying the Hockey Stick Identity**:
Therefore, we can apply the Hockey Stick Identity directly:
\[ \sum_{j=r}^{n} \binom{j}{r} = \binom{n+1}{r+1} \]

4. **Final Result**:
Thus, we conclude that:
\[ \sum_{i=0}^{n} \binom{n-i}{r} = \binom{n+1}{r+1} \]

### Conclusion:
The value of \( \sum_{i=0}^{n} \binom{n-i}{r} \) is equal to \( \binom{n+1}{r+1} \).