Let `T_(n)=(1)/((sqrt(n)+sqrt(n+1))(4sqrt(n)+4sqrt(n+1)))` and `S_(n)=sum_(r=1)^(n)T_(r)` then find `S_(15)`.

Explanation

To solve the problem, we need to find \( S_{15} \) where \( S_n = \sum_{r=1}^{n} T_r \) and \( T_n = \frac{1}{(\sqrt{n} + \sqrt{n+1})(4\sqrt{n} + 4\sqrt{n+1})} \).

### Step-by-Step Solution:

1. **Simplifying \( T_n \)**:
\[ T_n = \frac{1}{(\sqrt{n} + \sqrt{n+1})(4\sqrt{n} + 4\sqrt{n+1})} \]
We can factor out the constants in the denominator:
\[ T_n = \frac{1}{4(\sqrt{n} + \sqrt{n+1})^2} \]

2. **Rationalizing the Denominator**:
To simplify further, we can rationalize \( T_n \):
\[ T_n = \frac{1}{4(\sqrt{n} + \sqrt{n+1})^2} = \frac{(\sqrt{n+1} - \sqrt{n})}{4(n+1 - n)} = \frac{\sqrt{n+1} - \sqrt{n}}{4} \]

3. **Expressing \( S_n \)**:
Now, we can express \( S_n \):
\[ S_n = \sum_{r=1}^{n} T_r = \sum_{r=1}^{n} \frac{\sqrt{r+1} - \sqrt{r}}{4} \]
This is a telescoping series.

4. **Evaluating the Telescoping Series**:
The sum simplifies as follows:
\[ S_n = \frac{1}{4} \left( (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \ldots + (\sqrt{n+1} - \sqrt{n}) \right) \]
Most terms cancel out:
\[ S_n = \frac{1}{4} (\sqrt{n+1} - \sqrt{1}) = \frac{\sqrt{n+1} - 1}{4} \]

5. **Finding \( S_{15} \)**:
Now we can find \( S_{15} \):
\[ S_{15} = \frac{\sqrt{15+1} - 1}{4} = \frac{\sqrt{16} - 1}{4} = \frac{4 - 1}{4} = \frac{3}{4} \]

### Final Answer:
\[ S_{15} = \frac{3}{4} \]