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- Apr 1st 2012, 03:10 AM #1

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- Apr 1st 2012, 03:17 AM #2
## Re: The value of x

Wolfram didn't give an exact answer, so I doubt there is one...

sum 1/(Sqrt[2n] + Sqrt[2*(n+1)] ) with n from 0 to 1003 - Wolfram|Alpha

- Apr 1st 2012, 03:24 AM #3

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- Apr 1st 2012, 03:37 AM #4

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- Apr 1st 2012, 07:01 AM #5

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- Apr 1st 2012, 08:22 PM #6

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- Apr 2nd 2012, 08:16 AM #7

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- Apr 9th 2012, 06:43 PM #8

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## Re: The value of x

I'm surprised no one was able to figure this out until now (including me). It's actually fairly simple.

$\displaystyle x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \frac{1}{\sqrt n +\sqrt{n+1}} $

$\displaystyle x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \frac{\sqrt n -\sqrt{n+1}}{(\sqrt n -\sqrt{n+1})(\sqrt n +\sqrt{n+1})}$

$\displaystyle x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \sqrt{n+1} -\sqrt n$

Now it is a telescoping series.

$\displaystyle x=\frac{1}{\sqrt 2}(\sqrt{1003+1} -\sqrt 0)$

$\displaystyle x= \sqrt{502}$