Calculating x in Equation (1): A Telescopic Sum

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The value of x in the telescopic sum defined by the equation $\displaystyle x=\frac{1}{\sqrt{0}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{4}}+...+\frac{1}{\sqrt{2006}+\sqrt{2008}}$ is calculated to be $\frac{\sqrt{2008}}{2}$, which equals approximately 22.405356502. The telescopic nature of the sum simplifies the calculation, allowing for the cancellation of intermediate terms. The final result can also be expressed as $\sqrt{502}$.

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chisigma
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From mathhelpforum.com...

$\displaystyle x=\frac{1}{\sqrt{0}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{4}}+...+\frac{1}{\sqrt{2006}+\sqrt{2008}}$ (1)

... what is the value of x?...

The (1) is a telescopic sum...

$\displaystyle x=\frac{\sqrt{2}}{2} + \frac{\sqrt{4}-\sqrt{2}}{2} + \frac{\sqrt{6}-\sqrt{4}}{2}+... + \frac{\sqrt{2008}-\sqrt{2006}}{2}=\frac{\sqrt{2008}}{2}=22.405356502...$ (2)

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$\chi$ $\sigma$
 
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Correct.

Of course, ##\dfrac{\sqrt{2008}}{2}=\sqrt{502} ## .
 
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