Find the sum `(1)/(1+sqrt(2))+(1)/(sqrt(2)+sqrt(3))+(1)/(sqrt(3)+sqrt(4))+..........` upto 99 terms.

Answer

To find the sum

\[ S = \frac{1}{1+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + \frac{1}{\sqrt{3}+\sqrt{4}} + \ldots \]

up to 99 terms, we can start by rewriting each term in a more manageable form.

### Step 1: Rewrite Each Term
We can rewrite each term in the series using the identity for the difference of squares. Specifically, we can multiply the numerator and denominator of each term by the conjugate of the denominator.

For the first term:

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

For the second term:

\[ \frac{1}{\sqrt{2}+\sqrt{3}} \cdot \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}} = \frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})} = \frac{\sqrt{2}-\sqrt{3}}{2 - 3} = \sqrt{3} - \sqrt{2} \]

For the third term:

\[ \frac{1}{\sqrt{3}+\sqrt{4}} \cdot \frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}} = \frac{\sqrt{3}-\sqrt{4}}{(\sqrt{3}+\sqrt{4})(\sqrt{3}-\sqrt{4})} = \frac{\sqrt{3}-\sqrt{4}}{3 - 4} = \sqrt{4} - \sqrt{3} \]

Continuing this way, we can see that the general term can be expressed as:

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

### Step 2: Write the Sum
Thus, the sum \( S \) can be rewritten as:

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

### Step 3: Simplify the Sum
Notice that this is a telescoping series. Most terms will cancel out:

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

### Final Answer
Thus, the sum of the series up to 99 terms is:

\[ \boxed{9} \]