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

Options

A. `2.^(n)C_(r+2)`
B. `.^(n+1)C_(r+1)`
C. `.^(n+2)C_(r+2)`
D. None of these

Answer

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

### Step-by-step Solution:

1. **Write the Expression**:
We start with the expression:
\[ \binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2} \]

2. **Rearranging the Terms**:
We can rearrange the terms to group the coefficients:
\[ \binom{n}{r} + \binom{n}{r-1} + \binom{n}{r-1} + \binom{n}{r-2} \]
Here, we have grouped \( 2\binom{n}{r-1} \) as \( \binom{n}{r-1} + \binom{n}{r-1} \).

3. **Using the Identity**:
We can apply the identity:
\[ \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r} \]
to the first two terms:
\[ \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r} \]

4. **Applying the Identity Again**:
Now we apply the identity again to the next two terms:
\[ \binom{n}{r-1} + \binom{n}{r-2} = \binom{n+1}{r-1} \]

5. **Combining the Results**:
Now we can combine the results:
\[ \binom{n+1}{r} + \binom{n+1}{r-1} \]
Again, we apply the identity:
\[ \binom{n+1}{r} + \binom{n+1}{r-1} = \binom{n+2}{r} \]

6. **Final Result**:
Therefore, we conclude that:
\[ \binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2} = \binom{n+2}{r} \]

### Final Answer:
\[ \binom{n+2}{r} \]