`sum_(k=1)^(360)((1)/(ksqrt(k+1)+(k+1)sqrt(k)))` is the ratio of two relative prime positive integers m and n. The value of `(m+n)` is equal to

Options

A. 43
B. 41
C. 39
D. 37

Answer

To solve the problem, we need to evaluate the sum:

\[ S = \sum_{k=1}^{360} \frac{1}{k \sqrt{k+1} + (k+1) \sqrt{k}} \]

### Step 1: Simplify the Denominator
We start by factoring out \(\sqrt{k(k+1)}\) from the denominator:

\[ S = \sum_{k=1}^{360} \frac{1}{\sqrt{k(k+1)} \left(\sqrt{k} + \sqrt{k+1}\right)} \]

### Step 2: Rationalize the Denominator
Next, we rationalize the denominator by multiplying the numerator and denominator by \(\sqrt{k+1} - \sqrt{k}\):

\[ S = \sum_{k=1}^{360} \frac{\sqrt{k+1} - \sqrt{k}}{k(k+1) \left(k+1 - k\right)} \]

This simplifies to:

\[ S = \sum_{k=1}^{360} \frac{\sqrt{k+1} - \sqrt{k}}{k(k+1)(\sqrt{k+1} - \sqrt{k})} \]

### Step 3: Simplify Further
The denominator simplifies to:

\[ S = \sum_{k=1}^{360} \frac{\sqrt{k+1} - \sqrt{k}}{k(k+1)(\sqrt{k+1} - \sqrt{k})} = \sum_{k=1}^{360} \frac{1}{\sqrt{k(k+1)}} \]

### Step 4: Telescoping Series
Recognizing that this forms a telescoping series:

\[ S = \sum_{k=1}^{360} \left( \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k+1}} \right) \]

### Step 5: Evaluate the Series
This telescopes down to:

\[ S = 1 - \frac{1}{\sqrt{361}} = 1 - \frac{1}{19} = \frac{18}{19} \]

### Step 6: Identify m and n
The result \(\frac{18}{19}\) is in the form \(\frac{m}{n}\) where \(m = 18\) and \(n = 19\). Since 18 and 19 are relatively prime, we can proceed to find \(m+n\).

### Step 7: Calculate m+n
Thus, we have:

\[ m+n = 18 + 19 = 37 \]

### Final Answer
The value of \(m+n\) is:

\[ \boxed{37} \]