`""^(n)C_(r+1)+^(n)C_(r-1)+2.""^(n)C_(r)=`

Options

A. `""^(n+2)C_(r+1)`
B. `""^(n+1)C_(r)`
C. `""^(n+1)C_(r+1)`
D. `""^(n+2)C_(r)`

Answer

To solve the expression \( \binom{n}{r+1} + \binom{n}{r-1} + 2 \binom{n}{r} \), we will use the properties of binomial coefficients.

### Step-by-Step Solution:

1. **Write down the expression**:
\[ \binom{n}{r+1} + \binom{n}{r-1} + 2 \binom{n}{r} \]

2. **Apply the Pascal's Identity**:
Recall that:
\[ \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r} \]
Using this identity, we can combine \( \binom{n}{r} \) and \( \binom{n}{r-1} \):
\[ \binom{n}{r-1} + \binom{n}{r} = \binom{n+1}{r} \]
Thus, we can rewrite the expression as:
\[ \binom{n}{r+1} + \binom{n+1}{r} + \binom{n}{r} \]

3. **Rearranging the expression**:
Now, we can express \( \binom{n}{r+1} \) using the same identity:
\[ \binom{n}{r+1} = \binom{n}{r} + \binom{n-1}{r} \]
Substituting this back into our expression gives:
\[ \left( \binom{n}{r} + \binom{n-1}{r} \right) + \binom{n+1}{r} + \binom{n}{r} \]
This simplifies to:
\[ 2 \binom{n}{r} + \binom{n-1}{r} + \binom{n+1}{r} \]

4. **Using Pascal's Identity again**:
Now, we can again apply Pascal's identity:
\[ \binom{n-1}{r} + \binom{n}{r} = \binom{n}{r+1} \]
Therefore, we can write:
\[ \binom{n+1}{r+1} + \binom{n}{r} \]

5. **Final expression**:
Putting it all together, we find:
\[ \binom{n+1}{r+1} + \binom{n+1}{r} = \binom{n+2}{r+1} \]

Thus, the final result is:
\[ \binom{n+2}{r+1} \]