If `sum_(k = 1)^(oo) (1)/((k + 2)sqrt(k) + ksqrt(k + 2)) = (sqrt(a) + sqrt(b))/(sqrt(c))`, where `a, b, c in N` and `a,b,c in [1, 15]`, then a + b + c is equal to

Answer

To solve the problem, we need to evaluate the infinite series:

\[ \sum_{k=1}^{\infty} \frac{1}{(k + 2)\sqrt{k} + k\sqrt{k + 2}} \]

### Step 1: Simplify the Denominator
We start by factoring out \(\sqrt{k + 2}\) and \(\sqrt{k}\) from the denominator:

\[ (k + 2)\sqrt{k} + k\sqrt{k + 2} = \sqrt{k + 2}(\sqrt{k + 2} + \sqrt{k}) + \sqrt{k}(\sqrt{k + 2} + \sqrt{k}) \]

This can be rewritten as:

\[ \sqrt{k}\sqrt{k + 2}(\sqrt{k + 2} + \sqrt{k}) \]

### Step 2: Multiply by the Conjugate
Next, we multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{\sqrt{k + 2} - \sqrt{k}}{(k + 2) - k} = \frac{\sqrt{k + 2} - \sqrt{k}}{2} \]

This gives us:

\[ \sum_{k=1}^{\infty} \frac{\sqrt{k + 2} - \sqrt{k}}{2} \]

### Step 3: Split the Series
Now, we can split the series into two parts:

\[ \sum_{k=1}^{\infty} \frac{1}{2} \left( \sqrt{k + 2} - \sqrt{k} \right) \]

This can be rewritten as:

\[ \frac{1}{2} \left( \sum_{k=1}^{\infty} \sqrt{k + 2} - \sum_{k=1}^{\infty} \sqrt{k} \right) \]

### Step 4: Telescoping Series
Notice that this is a telescoping series. Most terms will cancel out:

\[ \frac{1}{2} \left( \sqrt{3} - \sqrt{1} + \sqrt{4} - \sqrt{2} + \sqrt{5} - \sqrt{3} + \ldots \right) \]

### Step 5: Evaluate the Remaining Terms
After cancellation, we are left with:

\[ \frac{1}{2} \left( \sqrt{2} + 1 \right) \]

### Step 6: Final Expression
Now, we can express this in the required form:

\[ \frac{\sqrt{2} + 1}{2} \]

### Step 7: Identify \(a\), \(b\), and \(c\)
We can match this with the form \(\frac{\sqrt{a} + \sqrt{b}}{\sqrt{c}}\):

- \(a = 2\)
- \(b = 1\)
- \(c = 4\)

### Step 8: Calculate \(a + b + c\)
Finally, we calculate:

\[ a + b + c = 2 + 1 + 4 = 7 \]

### Conclusion
Thus, the answer is:

\[ \boxed{7} \]