\( S = \frac { 1 } { \sqrt { 2 } + 1 } + \frac { 1 } { \sqrt { 3 } + \sqrt { 2 } } + \frac { 1 } { \sqrt { 4 } + \sqrt { 3 } } + \ldots + \frac { 1 } { \sqrt { 100 } + \sqrt { 99 } } \)

To simplify the series \( S = \frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{3} + \sqrt{2}} + \frac{1}{\sqrt{4} + \sqrt{3}} + \ldots + \frac{1}{\sqrt{100} + \sqrt{99}} \), we can use the technique of rationalizing the denominator for each term. For each term in the series, we can write: \[ \frac{1}{\sqrt{n} + \sqrt{n-1}} = \frac{\sqrt{n} - \sqrt{n-1}}{(\sqrt{n} + \sqrt{n-1})(\sqrt{n} - \sqrt{n-1})} = \frac{\sqrt{n} - \sqrt{n-1}}{n - (n-1)} = \sqrt{n} - \sqrt{n-1} \] Therefore, the series can be rewritten as: \[ S = (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \ldots + (\sqrt{100} - \sqrt{99}) \] If we look closely, this is a telescoping series. Most terms will cancel out: \[ S = \sqrt{100} - 1 = 10 - 1 = 9 \] Thus, the value of the series \( S \) is: \[ \boxed{9} \]