(1) \ {1 \over {\sqrt{100} + \sqrt{99}}} + {1 \over {\sqrt{99} + \sqrt{98}}} + {1 \over {\sqrt{98} + \sqrt{97}}} + {1 \over {\sqrt{97} + \sqrt{96}}} ... +{1 \over {\sqrt{2} + \sqrt{1}}} = ?
因為
\begin{align*}
{1 \over { \sqrt{100} + \sqrt{99}}} &= {1 \over { \sqrt{100} + \sqrt{99}}} \times { { \sqrt{100} - \sqrt{99}} \over { \sqrt{100}} - \sqrt{99}} \\
&= {\sqrt{100} - \sqrt{99} \over (\sqrt{100})^2 - (\sqrt{99})^2} \\
&= {{\sqrt{100} - \sqrt{99}} \over 100 - 99}\\
&= {\sqrt{100} - \sqrt{99}} \\
\end{align*}
所以
{1 \over {\sqrt{100} + \sqrt{99}}} + {1 \over {\sqrt{99} + \sqrt{98}}} + {1 \over {\sqrt{98} + \sqrt{97}}} + {1 \over {\sqrt{97} + \sqrt{96}}} ... +{1 \over {\sqrt{2} + \sqrt{1}}} =\\
{\sqrt{100} - \sqrt{99}} + {\sqrt{99} - \sqrt{98}} + {\sqrt{98} - \sqrt{97}} + {\sqrt{97} - \sqrt{96}} ... + {\sqrt{2} - \sqrt{1}} =\\
{\sqrt{100} - \sqrt{1}} = 9
(2) \ {1 \over {\sqrt{100} - \sqrt{99}}} - {1 \over {\sqrt{99} - \sqrt{98}}} + {1 \over {\sqrt{98} - \sqrt{97}}} - {1 \over {\sqrt{97} - \sqrt{96}}} ... +{1 \over {\sqrt{2} - \sqrt{1}}} = ?
因為
\begin{align*}
{1 \over { \sqrt{100} - \sqrt{99}}} &= {1 \over { \sqrt{100} - \sqrt{99}}} \times { { \sqrt{100} + \sqrt{99}} \over { \sqrt{100}} + \sqrt{99}} \\
&= {\sqrt{100} + \sqrt{99} \over (\sqrt{100})^2 - (\sqrt{99})^2} \\
&= {{\sqrt{100} + \sqrt{99}} \over 100 - 99}\\
&= {\sqrt{100} + \sqrt{99}} \\
\end{align*}
所以
{1 \over {\sqrt{100} - \sqrt{99}}} - {1 \over {\sqrt{99} - \sqrt{98}}} + {1 \over {\sqrt{98} - \sqrt{97}}} - {1 \over {\sqrt{97} - \sqrt{96}}} ... +{1 \over {\sqrt{2} - \sqrt{1}}} =\\
({\sqrt{100} + \sqrt{99}}) - ({\sqrt{99} + \sqrt{98}}) + ({\sqrt{98} + \sqrt{97}}) - ({\sqrt{97} + \sqrt{96}}) ... + ({\sqrt{2} + \sqrt{1}}) =\\
{\sqrt{100} + \sqrt{99}} - {\sqrt{99} - \sqrt{98}} + {\sqrt{98} + \sqrt{97}} - {\sqrt{97} - \sqrt{96}} ... + {\sqrt{2} + \sqrt{1}} = \\
{\sqrt{100} + \sqrt{1}} = 11
(3) \ {1 \over {100 \sqrt{99} + 99 \sqrt{100}}} + {1 \over {99 \sqrt{98} + 98\sqrt{99}}} + ... + {1 \over {3 \sqrt 2} + 2 \sqrt 3} + {1 \over {2 \sqrt 1 + \sqrt 2}} = ?
因為
\begin{align*}
{1 \over {100 \sqrt{99} + 99 \sqrt{100}}} &= {1 \over {\sqrt{100}^2 \sqrt{99} + \sqrt{99}^2 \sqrt{100}}} \\
&= {1 \over {\sqrt{100} \sqrt{99} \times ( \sqrt{100} + \sqrt{99} ) }} \\
&= {1 \over {\sqrt{100} \sqrt{99} \times ( \sqrt{100} + \sqrt{99} ) }} \ \times { \sqrt{100} - \sqrt{99} \over \sqrt{100} - \sqrt{99} } \\
&= {\sqrt{100} - \sqrt{99} \over {\sqrt{100} \sqrt{99} \times ( \sqrt{100}^2 - \sqrt{99}^2 ) }} \\
&= {\sqrt{100} - \sqrt{99} \over {\sqrt{100} \sqrt{99} }} \\
&= {1 \over \sqrt{99}} - {1 \over \sqrt{100}} \\
\end{align*}
所以
{1 \over {100 \sqrt{99} + 99 \sqrt{100}}} + {1 \over {99 \sqrt{98} + 98\sqrt{99}}} + ... + {1 \over {3 \sqrt 2} + 2 \sqrt 3} + {1 \over {2 \sqrt 1 + \sqrt 2}} = \\
({1 \over \sqrt{99}} - {1 \over \sqrt{100}}) + ({1 \over \sqrt{98}} - {1 \over \sqrt{99}}) + ... + ({1 \over \sqrt{2}} - {1 \over \sqrt{3}}) + ({1 \over \sqrt{1}} - {1 \over \sqrt{2}}) = \\
- {1 \over \sqrt{100}} + {1 \over \sqrt{1}} = -{1 \over 10} + 1 = {9 \over 10}
練習題：
(4) \ {1 \over {100\sqrt{99} - 99\sqrt{100}}} - {1 \over {99\sqrt{98} - 98\sqrt{99}}} + {1 \over {98\sqrt{97} - 97\sqrt{98}}} - ... +{1 \over {2\sqrt{1} - \sqrt{2}}} = ?