`""^(n)C_(r)+2""^(n)C_(r-1)+^(n)C_(r-2)` is equal to

Options

A. `""^(n+1)C_(r)`
B. `""^(n)C_(r+1)`
C. `""^(n-1)C_(r+1)`
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 combinations. Here’s a step-by-step breakdown:

### Step 1: Rewrite the expression
We start with the expression:
\[ \binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2} \]

### Step 2: Use the property of combinations
We know that:
\[ \binom{n}{r-1} + \binom{n}{r-2} = \binom{n+1}{r-1} \]
This is derived from the identity:
\[ \binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k} \]

### Step 3: Rewrite the expression using the identity
Now, we can rewrite \( 2\binom{n}{r-1} \) as:
\[ \binom{n}{r-1} + \binom{n}{r-1} = \binom{n+1}{r} \]
Thus, we can combine the terms:
\[ \binom{n}{r} + \binom{n}{r-1} + \binom{n}{r-1} + \binom{n}{r-2} = \binom{n}{r} + \left(\binom{n}{r-1} + \binom{n}{r-2}\right) + \binom{n}{r-1} \]

### Step 4: Combine the terms
Now we can combine the terms:
\[ \binom{n}{r} + \binom{n+1}{r-1} = \binom{n+1}{r} \]
Thus, we have:
\[ \binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2} = \binom{n+2}{r} \]

### Final Result
Therefore, the expression simplifies to:
\[ \binom{n+2}{r} \]