​Simplify the following.

$$\frac{1}{\sqrt2+\sqrt1}+\frac{1}{\sqrt3+\sqrt2}+\frac{1}{\sqrt4+\sqrt3}+...+\frac{1}{\sqrt{100}+\sqrt{99}}$$​​

Correct option is C

Given:
​$$\frac{1}{\sqrt2+\sqrt1}+\frac{1}{\sqrt3+\sqrt2}+\frac{1}{\sqrt4+\sqrt3}+...+\frac{1}{\sqrt{100}+\sqrt{99}}$$
Concept Used:
For each term of the form $$\frac{1}{\sqrt{n+1} + \sqrt{n}}$$,  we rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator  $${\sqrt{n+1} - \sqrt{n}}$$​
​$$\frac{1}{\sqrt{n+1} + \sqrt{n}} \times \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} = \sqrt{n+1} - \sqrt{n}$$​​
​Solution:​

Add all of the above given results:
​$$\sqrt{2} - \sqrt{1} + \sqrt{3} - \sqrt{2} + \sqrt{4} - \sqrt{3} + \dots + \sqrt{100} - \sqrt{99}\\\sqrt{100} - \sqrt{1} = 10 - 1 = 9\\$$​​
Thus, the simplified value of the sum is 9.