The value of $$\frac{1}{{1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + \frac{1}{{\sqrt 4 + \sqrt 5 }} + \frac{1}{{\sqrt 5 + \sqrt 6 }} + \frac{1}{{\sqrt 6 + \sqrt 7 }} + \frac{1}{{\sqrt 7 + \sqrt 8 }} + \frac{1}{{\sqrt 8 + \sqrt 9 }}{\text{is:}}$$

Solution (By Examveda Team)

$$\eqalign{ & \frac{1}{{1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + \frac{1}{{\sqrt 4 + \sqrt 5 }} + \frac{1}{{\sqrt 5 + \sqrt 6 }} + \frac{1}{{\sqrt 6 + \sqrt 7 }} + \frac{1}{{\sqrt 7 + \sqrt 8 }} + \frac{1}{{\sqrt 8 + \sqrt 9 }} \cr & = \frac{1}{{\sqrt 2 + 1}} + \frac{1}{{\sqrt 3 + \sqrt 2 }} + \frac{1}{{\sqrt 4 + \sqrt 3 }} + \frac{1}{{\sqrt 5 + \sqrt 4 }} + \frac{1}{{\sqrt 6 + \sqrt 5 }} + \frac{1}{{\sqrt 7 + \sqrt 6 }} + \frac{1}{{\sqrt 8 + \sqrt 7 }} + \frac{1}{{\sqrt 9 + \sqrt 8 }} \cr & {\text{After Rationalizing}} \cr & = \left( {\sqrt 2 - 1} \right) + \left( {\sqrt 3 - \sqrt 2 } \right) + \left( {\sqrt 4 - \sqrt 3 } \right) + \left( {\sqrt 5 - \sqrt 4 } \right) + \left( {\sqrt 6 - \sqrt 5 } \right) + \left( {\sqrt 7 - \sqrt 6 } \right) + \left( {\sqrt 8 - \sqrt 7 } \right) + \left( {\sqrt 9 - \sqrt 8 } \right) \cr & = \sqrt 9 - 1 \cr & = 3 - 1 \cr & = 2 \cr} $$