`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 given problem, we need to evaluate the summation:

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

We can break down the fraction \(\frac{r+2}{r+1}\) into two parts:

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

Thus, we can rewrite the summation as:

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

This can be split into two separate summations:

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

### Step 1: Evaluate the first summation

The first summation is:

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

This is a well-known result from the binomial theorem.

### Step 2: Evaluate the second summation

The second summation is:

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

We can relate this summation to the binomial coefficients by using 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 combinatorial interpretation of binomial coefficients.

### Step 3: Combine the results

Now, we can combine both results:

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

### Final Result

Thus, the final result of the summation is:

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