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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

Adola
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2 Answers2

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Hint:

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

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