This Spiral is So Radical Dude!

Well, this was a very simple telescoping sum. Firstly, we need to find where the particle tends to. We will separate the coordinate to 2, the x-coordinate and the y-coordinate. We will solve the x-coordinate first.

East is the direction of the positive x and west is the direction of the negative x. We have the x-coordinate equal to

$\dfrac{1}{\sqrt{1}+\sqrt{5}} - \dfrac{1}{\sqrt{3}+\sqrt{7}} + \dfrac{1}{\sqrt{5}+\sqrt{9}} - \ldots$

Rationalising the denominator for all terms we have

$\dfrac{\sqrt{5}-\sqrt{1}}{4}-\dfrac{\sqrt{7}-\sqrt{3}}{4}+\dfrac{\sqrt{9}-\sqrt{5}}{4}-\dfrac{\sqrt{11}-\sqrt{7}}{4}+\ldots$

$\dfrac{(\sqrt{5}-\sqrt{1})-(\sqrt{7}-\sqrt{3})+(\sqrt{9}-\sqrt{5})-(\sqrt{11}-\sqrt{7}) + \ldots}{4}$

Solving, we have

$\dfrac{\sqrt{5}-\sqrt{1}-\sqrt{7}+\sqrt{3}+\sqrt{9}-\sqrt{5}-\sqrt{11}+\sqrt{7}+\ldots}{4}$

In this expression, all terms except $-\sqrt{1}$ and $\sqrt{3}$ can be grouped with its negative. Thus, this expression simplifies to

$\dfrac{\sqrt{3}-\sqrt{1}}{4}$

Thus, the x-coordinate is $\dfrac{\sqrt{3}-1}{4}$ .

Using the same method, we determine that the value of the y-coordinate is $\dfrac{2-\sqrt{2}}{4}$

Now, to determine the magnitude, we will apply $\sqrt{x^2 + y^2}$ . At this point, we can use a calculator to determine that the magnitude is 0.23439... By multiplying by 1000 and inputting its floor function, we get an answer of 234.