Value of `1/(sqrt2+1)+1/(sqrt3+sqrt2)+1/(sqrt4+sqrt3)+....+1/(sqrt100+sqrt99)` is

Options

A. 1
B. 9
C. `sqrt99`
D. `sqrt(99-1)`

Answer

To find the value of 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 will rationalize each term in the series.

### Step-by-Step Solution:

1. **Rationalize the first term**:
\[ \frac{1}{\sqrt{2} + 1} \cdot \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{(\sqrt{2})^2 - 1^2} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1 \]

2. **Rationalize the second term**:
\[ \frac{1}{\sqrt{3} + \sqrt{2}} \cdot \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2} \]

3. **Rationalize the third term**:
\[ \frac{1}{\sqrt{4} + \sqrt{3}} \cdot \frac{\sqrt{4} - \sqrt{3}}{\sqrt{4} - \sqrt{3}} = \frac{\sqrt{4} - \sqrt{3}}{(\sqrt{4})^2 - (\sqrt{3})^2} = \frac{\sqrt{4} - \sqrt{3}}{4 - 3} = \sqrt{4} - \sqrt{3} = 2 - \sqrt{3} \]

4. **Continue rationalizing subsequent terms**:
- For \( n = 4 \):
\[ \frac{1}{\sqrt{5} + \sqrt{4}} \cdot \frac{\sqrt{5} - \sqrt{4}}{\sqrt{5} - \sqrt{4}} = \sqrt{5} - 2 \]
- For \( n = 5 \):
\[ \frac{1}{\sqrt{6} + \sqrt{5}} \cdot \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} - \sqrt{5}} = \sqrt{6} - \sqrt{5} \]
- Continue this process up to \( n = 100 \).

5. **Combine all terms**:
The series can be expressed as:
\[ S = (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (2 - \sqrt{3}) + (\sqrt{5} - 2) + \ldots + (\sqrt{100} - \sqrt{99}) \]

6. **Notice the cancellation**:
When we add all these terms, we see that most terms cancel out:
- \( -1 \) from the first term remains.
- \( +\sqrt{2} \) from the first term cancels with \( -\sqrt{2} \) from the second term.
- \( +\sqrt{3} \) from the second term cancels with \( -\sqrt{3} \) from the third term.
- This pattern continues until \( +\sqrt{99} \) cancels with \( -\sqrt{99} \) from the last term.

7. **Final result**:
The only terms that do not cancel are \( -1 \) from the first term and \( +\sqrt{100} \) from the last term:
\[ S = \sqrt{100} - 1 = 10 - 1 = 9 \]

### Final Answer:
The value of the series is \( \boxed{9} \).