$\displaystyle\sum_{i=0}^n \frac{2^i}{1+x^{2^{i}}}$
What technique is applicable here? I can't find a way to manipulate this sum to make it telescope. Just guide me.
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$$\frac{2^{i+1}}{1-x^{2^{i+1}}} = \frac{2^{i+1}}{(1-x^{2^i})(1+x^{2^i})} = 2^i \left ( \frac1{1-x^{2^i}} + \frac1{1+x^{2^i}}\right )$$
Therefore
$$\frac{2^i}{1+x^{2^i}} = \frac{2^{i+1}}{1-x^{2^{i+1}}} - \frac{2^i}{1-x^{2^i}}$$
Can you take it from here?
Ron Gordon
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@AdityaParson: the trick is knowing that $$x^{2^{i+1}} = (x^2)^i$$ Then you somehow just know to exploit the fact that $$1+y = \frac{1-y^2}{1-y}$$ – Ron Gordon May 10 '14 at 13:01
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Nicely done. Can you give me some insight into your motivation to this approach? Please – Adienl May 10 '14 at 13:02
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@RonGordon Huh? $x^{2^{i+1}} = (x^{2^i})^2$, as you have used above. – Viktor Vaughn May 10 '14 at 13:59
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1@SpamIAm: Thanks for catching the typo. I meant $$x^{2^{i+1}} = (x^2)^{2^i}$$ – Ron Gordon May 10 '14 at 14:01