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

Options

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

Answer

To solve the problem \( \sum_{r=0}^{n} \frac{{n \choose r}}{r+2} \), we will follow a systematic approach using the properties of binomial coefficients and integration.

### Step-by-Step Solution

1. **Understanding the Binomial Coefficient**:
We know that the binomial theorem states:
\[ (1 + x)^n = \sum_{r=0}^{n} {n \choose r} x^r \]
This means that the sum of the binomial coefficients multiplied by \( x^r \) gives us \( (1+x)^n \).

**Hint**: Recall the binomial theorem and how it relates to binomial coefficients.

2. **Integrate the Binomial Expansion**:
To introduce the \( r+2 \) in the denominator, we can integrate the binomial expansion. We integrate both sides with respect to \( x \):
\[ \int (1+x)^n \, dx = \frac{(1+x)^{n+1}}{n+1} + C \]
Evaluating from 0 to \( x \):
\[ \int_0^x (1+t)^n \, dt = \frac{(1+x)^{n+1}}{n+1} - \frac{1}{n+1} \]

**Hint**: Remember that integrating a function can help introduce terms in the denominator.

3. **Relate the Integral to the Summation**:
The integral can also be expressed as:
\[ \int_0^x (1+t)^n \, dt = \sum_{r=0}^{n} {n \choose r} \frac{x^{r+1}}{r+1} \]
Setting \( x = 1 \):
\[ \sum_{r=0}^{n} {n \choose r} \frac{1}{r+1} = \frac{(1+1)^{n+1}}{n+1} - \frac{1}{n+1} = \frac{2^{n+1} - 1}{n+1} \]

**Hint**: Use the property of summation and integration to relate the two.

4. **Integrate Again**:
Now, we need to integrate again to bring in the \( r+2 \) term:
\[ \int_0^x (1+t)^{n+1} \, dt = \frac{(1+x)^{n+2}}{n+2} - \frac{1}{n+2} \]
This can be expressed as:
\[ \sum_{r=0}^{n} {n \choose r} \frac{x^{r+2}}{(r+1)(r+2)} \]

**Hint**: Integrating again helps to introduce higher powers in the denominator.

5. **Evaluate at \( x = 1 \)**:
Setting \( x = 1 \) gives:
\[ \sum_{r=0}^{n} {n \choose r} \frac{1}{(r+1)(r+2)} = \frac{(1+1)^{n+2}}{n+2} - \frac{1}{n+2} = \frac{2^{n+2} - 1}{n+2} \]

**Hint**: Evaluating at \( x = 1 \) simplifies the expression significantly.

6. **Final Expression**:
Therefore, we find that:
\[ \sum_{r=0}^{n} \frac{{n \choose r}}{r+2} = \frac{2^{n+2} - 1}{(n+1)(n+2)} \]

**Hint**: Ensure to simplify the final expression and check for any common factors.

### Conclusion
The final result for the summation \( \sum_{r=0}^{n} \frac{{n \choose r}}{r+2} \) is:
\[ \frac{2^{n+2} - 1}{(n+1)(n+2)} \]