# Thread: The value of x

1. ## The value of x

what is the value of x?

3. ## Re: The value of x

Originally Posted by Mhmh96

what is the value of x?
$x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \frac{1}{\sqrt n +\sqrt{n+1}}$

5. ## Re: 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)

6. ## Re: The value of x

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 $x = \sqrt{502}$

7. ## Re: 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}

8. ## 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.
$x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \frac{1}{\sqrt n +\sqrt{n+1}}$
$x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \frac{\sqrt n -\sqrt{n+1}}{(\sqrt n -\sqrt{n+1})(\sqrt n +\sqrt{n+1})}$
$x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \sqrt{n+1} -\sqrt n$
Now it is a telescoping series.
$x=\frac{1}{\sqrt 2}(\sqrt{1003+1} -\sqrt 0)$
$x= \sqrt{502}$