The value of x

**Mhmh96** · April 1st 2012, 03:10 AM

what is the value of x?

**Prove It** · April 1st 2012, 03:17 AM

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

**princeps** · April 1st 2012, 03:24 AM

Originally Posted by Mhmh96

what is the value of x?

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

**Mhmh96** · April 1st 2012, 03:37 AM

Helpful answers,thanks !

**Wilmer** · April 1st 2012, 07:01 AM

FOR k = 0 to 2006 STEP 2
x = x + 1 / [SQRT(k) + SQRT(k+2)]
NEXT k

x = 22.4053565...

SURPRISE(?): x^2 = 502 (integer)

**browni3141** · April 1st 2012, 08:22 PM

It appears that $\sum\frac{1}{\sqrt{2n}+\sqrt{2(n+1)}}$ from n = 0 to k is $\sqrt{(k+1)/2}$
It is almost certain that $x = $\sqrt{502}$$

**Wilmer** · April 2nd 2012, 08:16 AM

Well, for all the good this'll do:

1 / [SQRT(n) + SQRT(n + 2)] = SQRT{[n + 1 - SQRT(n^2 + 2n)] / 2}

**browni3141** · April 9th 2012, 06:43 PM

I'm surprised no one was able to figure this out until now (including me). It's actually fairly simple.
$x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \frac{1}{\sqrt n +\sqrt{n+1}}$
$x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \frac{\sqrt n -\sqrt{n+1}}{(\sqrt n -\sqrt{n+1})(\sqrt n +\sqrt{n+1})}$
$x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \sqrt{n+1} -\sqrt n$
Now it is a telescoping series.
$x=\frac{1}{\sqrt 2}(\sqrt{1003+1} -\sqrt 0)$
$x= \sqrt{502}$

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