Calculating square roots

$\begin{aligned} S & = \frac 1{1+\sqrt 2} + \frac 1{\sqrt 2 + \sqrt 3} + \frac 1{\sqrt 3 + \sqrt 4} + \cdots + \frac 1{\sqrt{63} + \sqrt {64}} \\ & = \frac {\sqrt 2 -1}{(\sqrt 2+1)(\sqrt 2-1)} + \frac {\sqrt 3 - \sqrt 2}{(\sqrt 3+\sqrt 2)(\sqrt 3-\sqrt 2)} + \frac {\sqrt 4 - \sqrt 3}{(\sqrt 4+\sqrt 3)(\sqrt 4-\sqrt 3)}+ \cdots + \frac {\sqrt{64} - \sqrt{63}}{(\sqrt{64}+\sqrt{63})(\sqrt{64}-\sqrt {63})} \\ & = \frac {\sqrt 2 -1}{2-1} + \frac {\sqrt 3 - \sqrt 2}{3-2} + \frac {\sqrt 4 - \sqrt 3}{4-3}+ \cdots + \frac {\sqrt{64} - \sqrt{63}}{64-63} \\ & = \blue{\cancel{\sqrt 2}} - 1 + \red{\cancel{\sqrt 3}} - \blue{\cancel{\sqrt 2}} + \blue{\cancel{\sqrt 4}} - \red{\cancel{\sqrt 3}} + \cdot + \sqrt {64} - \red{\cancel{\sqrt {63}}} \\ & = - 1 + \sqrt{64} = - 1 + 8 = \boxed 7 \end{aligned}$

$A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{63}+\sqrt{64}}$

$A=\frac{\sqrt{2}-1}{(1+\sqrt{2})(\sqrt{2}-1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})}+...+\frac{\sqrt{64}-\sqrt{63}}{(\sqrt{63}+\sqrt{64})(\sqrt{64}-\sqrt{63})}$

Using $(a-b)(a+b)=a^2-b^2$ , we have:

$A=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{64}-\sqrt{63}}{64-63}$

$\Leftrightarrow A=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{64}-\sqrt{63}$

$A=-1+\sqrt{64}=-1+8=\boxed{7}$