# Using $\pi$ to calculate square roots

Does anyone know why the following equation is true? $$\frac{(2d \pi^n+d^2) \sqrt{{2\pi^{2n}} + 2d\pi^n+d^2}} {2 \pi^{2n}+2d \pi^n+ \frac{3d^2}{4}} = d\sqrt{2} ,$$ as $n$ takes values from one to infinity.

-
I don't understand what's going on here. How is this different from your first question? Does Robert Israel's post answer either of the two questions? Why did you have to modify the question after accepting an answer? // It will be nice if you retain both the revisions of the question so that people can follow the changes made. – Srivatsan Dec 27 '11 at 2:24
I accepted Roberts opinion, the posting had an error. He did not answer my question. The truth is I am confused by your policies because each one of you has different interpretation as to what is an answer. You can delete the answer if you want. – Vassilis Parassidis Dec 27 '11 at 3:10

It isn't. You made some copying errors. Hint: $$\frac{\pi^{2n}}{2d} + \pi^n + \frac{d}{2} = \frac{(\pi^n + d)^2}{2d}$$
The identity has a little typo: the first term should be $\frac{\pi^{2n}}{2d}$. – Srivatsan Dec 26 '11 at 22:52