Value of $ \sum \limits_{k=1}^{81} \frac{1}{\sqrt{k} + \sqrt{k+1}} = \frac{1}{\sqrt{1} + \sqrt{2}} + \cdots + \frac{1}{\sqrt{80} + \sqrt{81}} $?
Solution 1:
Note that
$$ \frac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{n+1} - \sqrt{n} $$
Then you have a telescoping sum
Solution 2:
As \begin{align} \frac{1}{\sqrt{i}+\sqrt{i+1}} = \frac{\sqrt{i}-\sqrt{i+1}}{\sqrt{i}-\sqrt{i+1}} \cdot \frac{1}{\sqrt{i}+\sqrt{i+1}} = \frac{\sqrt{i}-\sqrt{i+1}}{-1} = {\sqrt{i+1}-\sqrt{i}} \end{align} Thus \begin{align} \sum_{i=1}^{80} \frac{1}{\sqrt{i}+\sqrt{i+1}} = \sum_{i=1}^{80} {\sqrt{i+1}-\sqrt{i}} = \sqrt{80+1} -\sqrt{1} = 8 \end{align}