The sum of the series frac{1}{x+1}+frac{2}{x^ | vedclass

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Question

(A) Let the sum be $S = \sum_{k=0}^{100} \frac{2^k}{x^{2^k}+1}$.We use the identity $\frac{1}{x-1} - \frac{1}{x+1} = \frac{2}{x^2-1}$.More generally,$

Vedclass · Accepted Answer

$1-\frac{2^{101}}{2^{2^{101}}-1}$

Vedclass · Answer

(A) Let the sum be $S = \sum_{k=0}^{100} \frac{2^k}{x^{2^k}+1}$.We use the identity $\frac{1}{x-1} - \frac{1}{x+1} = \frac{2}{x^2-1}$.More generally,$\frac{2^k}{x^{2^k}-1} - \frac{2^k}{x^{2^k}+1} = \frac{2^{k+1}}{x^{2^{k+1}}-1}$.Rearranging,$\frac{2^k}{x^{2^k}+1} = \frac{2^k}{x^{2^k}-1} - \frac{2^{k+1}}{x^{2^{k+1}}-1}$.This is a telescoping sum.$S = \left( \frac{1}{x-1} - \frac{2}{x^2-1} \right) + \left( \frac{2}{x^2-1} - \frac{4}{x^4-1} \right) + \ldots + \left( \frac{2^{100}}{x^{2^{100}}-1} - \frac{2^{101}}{x^{2^{101}}-1} \right)$.$S = \frac{1}{x-1} - \frac{2^{101}}{x^{2^{101}}-1}$.Given $x=2$,$S = \frac{1}{2-1} - \frac{2^{101}}{2^{2^{101}}-1} = 1 - \frac{2^{101}}{2^{2^{101}}-1}$.

The sum of the series $\frac{1}{x+1}+\frac{2}{x^{2}+1}+\frac{2^{2}}{x^{4}+1}+\ldots+\frac{2^{100}}{x^{2^{100}}+1}$ when $x=2$ is:

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