`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, we need to evaluate the summation:

\[ y = \sum_{r=0}^{n} \frac{r+2}{r+1} \binom{n}{r} \]

### Step 1: Rewrite the term in the summation
We can rewrite \(\frac{r+2}{r+1}\) as:

\[ \frac{r+2}{r+1} = \frac{r+1 + 1}{r+1} = 1 + \frac{1}{r+1} \]

Thus, we can express \(y\) as:

\[ y = \sum_{r=0}^{n} \left(1 + \frac{1}{r+1}\right) \binom{n}{r} \]

### Step 2: Separate the summation
Now, we can separate the summation into two parts:

\[ y = \sum_{r=0}^{n} \binom{n}{r} + \sum_{r=0}^{n} \frac{1}{r+1} \binom{n}{r} \]

### Step 3: Evaluate the first summation
The first summation is a well-known result:

\[ \sum_{r=0}^{n} \binom{n}{r} = 2^n \]

### Step 4: 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 is because \(\sum_{r=0}^{n} \binom{n+1}{r+1} = 2^{n+1}\).

### Step 5: Combine the results
Now we can combine both results:

\[ y = 2^n + \frac{1}{n+1} \cdot 2^{n+1} \]

### Step 6: Simplify the expression
We can factor out \(2^n\):

\[ y = 2^n + \frac{2^{n+1}}{n+1} = 2^n \left(1 + \frac{2}{n+1}\right) \]

### Final Result
Thus, the final expression for \(y\) is:

\[ y = 2^n \left(\frac{n+3}{n+1}\right) \]