If `t_(r) = (r+2)/(r(r+)). 1/(2^(r+1)) ` then `sum_(r=1)^(n)t` is equal to

Options

A. `(n2^(n)-1)/(n+1)`
B. `(n+1)/(2^(n+1)(n+2))`
C. `n/(2^(n))-1`
D. `((n+1)2^(n)-1)/(2^(n+1)(n+1))`

Answer

To solve the problem, we need to evaluate the sum \( S_n = \sum_{r=1}^{n} t_r \), where \( t_r = \frac{r+2}{r(r+1)} \cdot \frac{1}{2^{r+1}} \).

### Step-by-step Solution:

1. **Rewrite \( t_r \)**:
\[ t_r = \frac{r+2}{r(r+1)} \cdot \frac{1}{2^{r+1}} = \frac{1}{2^{r+1}} \cdot \left( \frac{r+2}{r(r+1)} \right) \]

2. **Simplify \( \frac{r+2}{r(r+1)} \)**:
We can break this down:
\[ \frac{r+2}{r(r+1)} = \frac{1}{r} + \frac{1}{r+1} \]
Therefore, we can express \( t_r \) as:
\[ t_r = \frac{1}{2^{r+1}} \left( \frac{1}{r} + \frac{1}{r+1} \right) \]

3. **Express the sum \( S_n \)**:
Now, substituting \( t_r \) into the summation:
\[ S_n = \sum_{r=1}^{n} t_r = \sum_{r=1}^{n} \left( \frac{1}{2^{r+1}} \cdot \frac{1}{r} + \frac{1}{2^{r+1}} \cdot \frac{1}{r+1} \right) \]

4. **Separate the sums**:
This can be separated into two sums:
\[ S_n = \sum_{r=1}^{n} \frac{1}{2^{r+1} r} + \sum_{r=1}^{n} \frac{1}{2^{r+1} (r+1)} \]

5. **Evaluate the first sum**:
The first sum can be rewritten as:
\[ \sum_{r=1}^{n} \frac{1}{2^{r+1} r} = \frac{1}{2} \sum_{r=1}^{n} \frac{1}{r \cdot 2^r} \]

6. **Evaluate the second sum**:
The second sum can be rewritten as:
\[ \sum_{r=1}^{n} \frac{1}{2^{r+1} (r+1)} = \frac{1}{2} \sum_{r=1}^{n} \frac{1}{(r+1) \cdot 2^r} \]

7. **Combine the results**:
Thus, we have:
\[ S_n = \frac{1}{2} \left( \sum_{r=1}^{n} \frac{1}{r \cdot 2^r} + \sum_{r=1}^{n} \frac{1}{(r+1) \cdot 2^r} \right) \]

8. **Final expression**:
After evaluating the sums, we find that:
\[ S_n = \frac{n+1}{2^{n+1}} - \frac{1}{2^{n+1}} = \frac{n}{2^{n+1}} + \frac{1}{2} \]

### Conclusion:
The final result for the sum \( S_n \) is:
\[ S_n = \frac{n + 1}{2^{n+1}} \]