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.
Using $\pi$ to calculate square roots
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algebra-precalculus
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0I 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. – 2011-12-27
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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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0I have edited the question above. – 2011-12-27