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Simplify $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \dots + \frac{1}{\sqrt{99} + \sqrt{100}}.$

Feb 15, 2020

#1
+24031
+2

Simplify
$$\dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + \dots + \dfrac{1}{\sqrt{99} + \sqrt{100}}$$.

$$\small{ \begin{array}{|rcll|} \hline &&\mathbf{\dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + \dots + \dfrac{1}{\sqrt{99} + \sqrt{100}}} \\\\ &=& \dfrac{1 - \sqrt{2}} {(1 + \sqrt{2})(1 - \sqrt{2})} + \dfrac{\sqrt{2} - \sqrt{3}} {(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3})} + \dfrac{\sqrt{3} - \sqrt{4}} {(\sqrt{3} + \sqrt{4})(\sqrt{3} - \sqrt{4})} + \dots + \dfrac{\sqrt{99} - \sqrt{100}} {(\sqrt{99} + \sqrt{100})(\sqrt{99} - \sqrt{100})} \\\\ &=& \dfrac{1 - \sqrt{2}} {1-2} + \dfrac{\sqrt{2} - \sqrt{3}} {2-3} + \dfrac{\sqrt{3} - \sqrt{4}} {3-4} + \dots + \dfrac{\sqrt{99} - \sqrt{100}} {99-100} \\\\ &=& \dfrac{1 - \sqrt{2}} {-1} + \dfrac{\sqrt{2} - \sqrt{3}} {-1} + \dfrac{\sqrt{3} - \sqrt{4}} {-1} + \dots + \dfrac{\sqrt{99} - \sqrt{100}} {-1} \\\\ &=& -1 +\underbrace{ \sqrt{2} -\sqrt{2} + \sqrt{3} -\sqrt{3} + \sqrt{4} + \dots -\sqrt{99}}_{=0} + \sqrt{100} \\\\ &=& -1 + \sqrt{100} \\\\ &=& -1 + 10 \\\\ &=& \mathbf{ 9 } \\ \hline \end{array} }$$

$$\dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + \dots + \dfrac{1}{\sqrt{99} + \sqrt{100}} = \mathbf{ 9}$$

Feb 15, 2020
edited by heureka  Feb 15, 2020
edited by heureka  Feb 15, 2020
#2
0

∑[1 / (sqrt(n) + sqrt(n+1)), n, 1, 99] = 9

Feb 15, 2020