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what is the value of x?

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- Apr 1st 2012, 03:10 AMMhmh96The value of x
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what is the value of x? - Apr 1st 2012, 03:17 AMProve ItRe: 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 AMprincepsRe: The value of x
- Apr 1st 2012, 03:37 AMMhmh96Re: The value of x
Helpful answers,thanks !

- Apr 1st 2012, 07:01 AMWilmerRe: The value of x
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) (Wondering) - Apr 1st 2012, 08:22 PMbrowni3141Re: The value of x
It appears that $\displaystyle $\sum\frac{1}{\sqrt{2n}+\sqrt{2(n+1)}}$$from n = 0 to k is $\displaystyle \sqrt{(k+1)/2}$

It is almost certain that $\displaystyle x = $\sqrt{502}$$ - Apr 2nd 2012, 08:16 AMWilmerRe: The value of x
Well, for all the good this'll do:

1 / [SQRT(n) + SQRT(n + 2)] = SQRT{[n + 1 - SQRT(n^2 + 2n)] / 2} - Apr 9th 2012, 06:43 PMbrowni3141Re: 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.

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

$\displaystyle x= \sqrt{502}$