Let ` S _(n ) = sum _( k =0) ^(n) (1)/( sqrt ( k +1) + sqrt k)` What is the value of `sum _( n =1) ^( 99) (1)/( S _(n ) + S _( n -1)) ?=`

Answer

To solve the problem, we will go through the steps systematically.

### Step 1: Define the sequence \( S_n \)

We are given:

\[ S_n = \sum_{k=0}^{n} \frac{1}{\sqrt{k+1} + \sqrt{k}} \]

### Step 2: Rationalize the denominator

To simplify \( S_n \), we can rationalize the denominator:

\[ \frac{1}{\sqrt{k+1} + \sqrt{k}} \cdot \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1} - \sqrt{k}} = \frac{\sqrt{k+1} - \sqrt{k}}{(\sqrt{k+1})^2 - (\sqrt{k})^2} \]

Using the identity \( a^2 - b^2 = (a+b)(a-b) \):

\[ = \frac{\sqrt{k+1} - \sqrt{k}}{(k+1) - k} = \sqrt{k+1} - \sqrt{k} \]

### Step 3: Rewrite \( S_n \)

Now, substituting back into the sum:

\[ S_n = \sum_{k=0}^{n} (\sqrt{k+1} - \sqrt{k}) \]

### Step 4: Evaluate the telescoping series

This is a telescoping series. When we expand it:

\[ S_n = (\sqrt{1} - \sqrt{0}) + (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \ldots + (\sqrt{n+1} - \sqrt{n}) \]

Most terms cancel out:

\[ S_n = \sqrt{n+1} - \sqrt{0} = \sqrt{n+1} \]

### Step 5: Find \( S_n + S_{n-1} \)

Now, we need to find \( S_n + S_{n-1} \):

\[ S_{n-1} = \sqrt{n} \]

Thus,

\[ S_n + S_{n-1} = \sqrt{n+1} + \sqrt{n} \]

### Step 6: Set up the main summation

We need to evaluate:

\[ \sum_{n=1}^{99} \frac{1}{S_n + S_{n-1}} = \sum_{n=1}^{99} \frac{1}{\sqrt{n+1} + \sqrt{n}} \]

### Step 7: Rationalize the denominator again

We can rationalize the denominator:

\[ \frac{1}{\sqrt{n+1} + \sqrt{n}} \cdot \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} = \frac{\sqrt{n+1} - \sqrt{n}}{(\sqrt{n+1})^2 - (\sqrt{n})^2} \]

This simplifies to:

\[ = \frac{\sqrt{n+1} - \sqrt{n}}{(n+1) - n} = \sqrt{n+1} - \sqrt{n} \]

### Step 8: Substitute back into the summation

Now substituting back into the summation:

\[ \sum_{n=1}^{99} (\sqrt{n+1} - \sqrt{n}) \]

### Step 9: Evaluate the telescoping series

This is again a telescoping series:

\[ = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \ldots + (\sqrt{100} - \sqrt{99}) \]

Most terms cancel out, leaving us with:

\[ = \sqrt{100} - \sqrt{1} = 10 - 1 = 9 \]

### Final Answer

Thus, the final value of the summation is:

\[ \boxed{9} \]