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}} $?

Question

I tried my best, but I am totally clueless about it. Worse thing is we were supposed to arrive at the answer in approximately $ 2 $ minutes. The correct answer is $ 8 $, right? Can you kindly explain how to arrive at it? I hope it won’t be too much bother. Thank you.

Answer 1

Note that

$$ \frac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{n+1} - \sqrt{n} $$

Then you have a telescoping sum

Answer 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}

Answer 3

$$\frac{\sqrt{2} - \sqrt{1} }{(2-1)} +\ldots +\frac{\sqrt{81} - \sqrt{80} }{(81-80)}=\sqrt{81}-\sqrt{1}=9-1=8 $$