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I have only an elementary knowledge of sequences and series so this is how I proceeded:
We first expand the sum and write down some of the initial terms $$ S = \sum_{i=0}^{n} {2^i \over x^{2^i} + 1} = \frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \ldots + \frac{2^n}{x^{2^n}+1} $$
Since every expression of the form $(x^k+1)$ is a factor of $(x^{2k} - 1)$ and $(x^k + 1)(x^k - 1) = (x^{2k} - 1)$;
We add and subtract the term $1 \over x-1$, and proceed by combining the terms with denominators of the above stated form
$$\begin{align} S &= \frac{1}{x-1} + - \frac{1}{x-1} + \frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \ldots + \frac{2^n}{x^{2^n}+1} \\ &= \frac{1}{x-1} + \frac{-2}{x^2-1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \ldots + \frac{2^n}{x^{2^n}+1} \\ &= \frac{1}{x-1} + \frac{-4}{x^4-1} + \frac{4}{x^4+1} + \ldots + \frac{2^n}{x^{2^n}+1} \\ &= \frac{1}{x-1} + \frac{-8}{x^8-1} + \frac{8}{x^8+1} + \ldots + \frac{2^n}{x^{2^n}+1} \\ \end{align}$$
Therefore, $$\begin{align} S &= \frac{1}{x-1} + \frac{-2^n}{x^{2^n}-1} + \frac{2^n}{x^{2^n}+1} \\ &= \frac{1}{x-1} + \frac{-2^{n+1}}{x^{2^{n+1}}-1} \\ \end{align}$$
Hence, the result
$$\sum_{i=0}^{n} {2^i \over x^{2^i} + 1} = \frac{1}{x-1} - \frac{2^{n+1}}{x^{2^{n+1}}-1} \\$$
- Is this result (and its derivation) correct?
- Can any further simplification be performed?
