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I have the following question:

Find the value of: $$ \frac {1} {\sqrt {4} + \sqrt {5}} + \frac {1} {\sqrt {5} + \sqrt {6}} + ... + \frac {1} {\sqrt {80} + \sqrt {81}}$$

The book already provide the answer for this question but it didn't include the ways. Hint or anything that will help is appreciated

  • 3
    Can you think of something that would make $\dfrac{1}{\sqrt{n} + \sqrt{n+1}}$ easier to handle? – 2017-01-19
  • 0
    What have you tried? Where did you get stuck? This question is missing lots of context. Please update it to show up your thoughts and attempt at the problem so we can more fully assist your learning as opposed to just doing your homework for you. – 2017-01-19
  • 0
    I didn't realize that you wanted someone to do your work for you. – 2017-01-19
  • 0
    Good edit. Why didn't you do the next step of collecting like terms? – 2017-01-19
  • 2
    [Rationalizing the denominator](http://math.stackexchange.com/a/60244/242) yields a telescoping sum. – 2017-01-19

2 Answers 2

8

Hint:

$$\frac1{\sqrt{n}+\sqrt{n+1}} = \sqrt{n+1}-\sqrt{n}$$

  • 0
    Another hint: "telescope". – 2017-01-19
4

$$\frac{\sqrt{4}-\sqrt{5}}{(\sqrt{4}+\sqrt{5})\cdot (\sqrt{4}-\sqrt{5})}+\frac{\sqrt{5}-\sqrt{6}}{(\sqrt{5}+\sqrt{6})\cdot (\sqrt{5}-\sqrt{6})}+ \ldots +\frac{\sqrt{80}-\sqrt{81}}{(\sqrt{80}+\sqrt{81})\cdot (\sqrt{80}-\sqrt{81})}= $$ $$=\frac{\sqrt{4}-\sqrt{5}+\sqrt{5}-\sqrt{6}+\ldots + \sqrt{80}-\sqrt{81}}{-1}=$$ $$=\frac{\sqrt{4}-\sqrt{81}}{-1}=$$ $$=\frac{2-9}{-1}=\frac{-7}{-1}=7$$