Find the value of $?$ in the given surds & fractions equation: $\frac{1}{\sqrt{2}+1} + \frac{1}{\sqrt{3}+\sqrt{2}} + \frac{1}{\sqrt{4}+\sqrt{3}} + \dots + \frac{1}{\sqrt{121}+\sqrt{120}} = ?$

Rationalizing each term of the form $\frac{1}{\sqrt{n+1}+\sqrt{n}}$: $\frac{\sqrt{n+1}-\sqrt{n}}{(n+1)-n} = \sqrt{n+1} - \sqrt{n}$. The series becomes: $(\sqrt{2}-1) + (\sqrt{3}-\sqrt{2}) + (\sqrt{4}-\sqrt{3}) + \dots + (\sqrt{121}-\sqrt{120})$. All intermediate terms cancel out, leaving: $\sqrt{121} - 1 = 11 - 1 = 10$.