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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.

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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

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It isn't. You made some copying errors. Hint: $$\frac{\pi^{2n}}{2d} + \pi^n + \frac{d}{2} = \frac{(\pi^n + d)^2}{2d}$$

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The identity has a little typo: the first term should be $\frac{\pi^{2n}}{2d}$. –  Srivatsan Dec 26 '11 at 22:52
    
You're right, I made a mistake. Can I withdraw the question? –  Vassilis Parassidis Dec 26 '11 at 23:00
    
@Vassili, you may want to increase your acceptance rate for your previous questions that have been answered. Questions are not "withdrawn" because they may prove to be useful to someone else in the future with a similar worry. –  Samuel Reid Dec 26 '11 at 23:08
    
I'm sorry for the mistake. I didn't pay enough attention. The question should read as follows –  Vassilis Parassidis Dec 27 '11 at 0:16
    
I have edited the question above. –  Vassilis Parassidis Dec 27 '11 at 0:46

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