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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

\[\sum_{n=1}^{9999}\frac{1}{(\sqrt{{n+1}}+\sqrt{n}\,\,)(\sqrt[4]{n+1}\,\,+\sqrt[4]{n}\,\,)}\]

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- Thread starter Albert
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- Thread starter
- #1

- Jan 25, 2013

- 1,225

\[\sum_{n=1}^{9999}\frac{1}{(\sqrt{{n+1}}+\sqrt{n}\,\,)(\sqrt[4]{n+1}\,\,+\sqrt[4]{n}\,\,)}\]

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- Feb 13, 2012

- 1,704

$\sum_{1}^{9999}\dfrac{1}{(\sqrt{{n+1}}+\sqrt{n}\,\,)(\sqrt[4]{n+1}\,\,+\sqrt[4]{n}\,\,)}$

$\displaystyle \frac{1}{(\sqrt{n+1} + \sqrt{n})\ (\sqrt[4]{n+1} + \sqrt[4]{n})} = \frac{(\sqrt{n+1} - \sqrt{n})\ (\sqrt[4] {n+1} - \sqrt[4] {n})}{\sqrt{n+1} - \sqrt{n}} = \sqrt[4] {n+1} - \sqrt[4] {n}$

... that is a 'telescopic sum'...

$\displaystyle S= \sqrt[4] {2} - \sqrt[4] {1} + \sqrt[4] {3} - \sqrt[4] {2} + ... + \sqrt[4] {10000} - \sqrt[4] {9999} = 10 - 1 = 9$

Kind regards

$\chi$ $\sigma$

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- Jan 25, 2013

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well done !