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 expression \(\binom{n-i}{r}\) represents the number of ways to choose \( r \) items from \( n-i \) items. As \( i \) varies from \( 0 \) to \( n \), \( n-i \) varies from \( n \) to \( 0 \).

2. **Rewriting the Summation**:
We can rewrite the summation by changing the index. Let \( j = n - i \). Then when \( i = 0 \), \( j = n \) and when \( i = n \), \( j = 0\). Thus, we can rewrite the summation as:
\[ \sum_{j=0}^{n} \binom{j}{r} \]

3. **Using the Hockey Stick Identity**:
The Hockey Stick Identity in combinatorics states that:
\[ \sum_{j=r}^{n} \binom{j}{r} = \binom{n+1}{r+1} \]
This means that the sum of combinations from \( r \) to \( n \) of choosing \( r \) items can be expressed as a single combination.

4. **Adjusting the Limits**:
Since our summation starts from \( j = 0 \) to \( n \), we can express it as:
\[ \sum_{j=0}^{n} \binom{j}{r} = \sum_{j=r}^{n} \binom{j}{r} + \sum_{j=0}^{r-1} \binom{j}{r} \]
The second part, \(\sum_{j=0}^{r-1} \binom{j}{r}\), is zero because we cannot choose \( r \) items from fewer than \( r \) items.

5. **Final Result**:
Therefore, we have:
\[ \sum_{j=0}^{n} \binom{j}{r} = \binom{n+1}{r+1} \]

Thus, the value of the original summation is:

\[ \sum_{i=0}^{n} \binom{n-i}{r} = \binom{n+1}{r+1} \]