`sum_(r=0)^(n)((r+2)/(r+1))*""^(n)C_(r)` is equal to :

Options

A. `(2^(n)(n+2)-1)/((n+1))`
B. `(2^(n)(n+1)-1)/((n+1))`
C. `(2^(n)(n+4)-1)/((n+1))`
D. `(2^(n)(n+3)-1)/((n+1))`

Answer

To solve the problem \( S_n = \sum_{r=0}^{n} \frac{r+2}{r+1} \binom{n}{r} \), we can break it down step by step.

### Step 1: Rewrite the expression
We start with the expression:
\[ S_n = \sum_{r=0}^{n} \frac{r+2}{r+1} \binom{n}{r} \]
We can separate the fraction:
\[ S_n = \sum_{r=0}^{n} \left( 1 + \frac{1}{r+1} \right) \binom{n}{r} \]
This simplifies to:
\[ S_n = \sum_{r=0}^{n} \binom{n}{r} + \sum_{r=0}^{n} \frac{1}{r+1} \binom{n}{r} \]

### Step 2: Evaluate the first summation
The first summation is simply:
\[ \sum_{r=0}^{n} \binom{n}{r} = 2^n \]

### Step 3: Evaluate the second summation
For the second summation, we can use the identity:
\[ \sum_{r=0}^{n} \frac{1}{r+1} \binom{n}{r} = \frac{1}{n+1} \sum_{r=0}^{n} \binom{n+1}{r+1} = \frac{1}{n+1} \cdot 2^{n+1} \]
This follows from the fact that \(\sum_{r=0}^{n} \binom{n+1}{r+1} = 2^{n+1}\).

### Step 4: Combine the results
Now, substituting back into our expression for \( S_n \):
\[ S_n = 2^n + \frac{1}{n+1} \cdot 2^{n+1} \]
We can factor out \( 2^n \):
\[ S_n = 2^n \left( 1 + \frac{2}{n+1} \right) \]
This simplifies to:
\[ S_n = 2^n \cdot \frac{n+2}{n+1} \]

### Final Result
Thus, the final result is:
\[ S_n = \frac{2^n (n + 2)}{n + 1} \]