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Differentiation can be used to find the sum of \( n \) terms or infinite terms in certain cases. e.g. we know that for \( |x|1 \)
\[
\begin{aligned}
\frac{1}{1-x} & =1+x+x^{2}+\ldots \\
\frac{1}{(1-x)^{2}} & =1+2 x+3 x^{2}+4 x^{3}+\ldots+n x^{n-1}+\ldots \\
& =\sum_{n=1}^{\infty} n x^{n-1} \\
\frac{2}{(1-x)^{3}} & =\sum_{n=2}^{\infty} n(n-1) x^{n-2}
\end{aligned}
\]
The sum of
\[
\frac{1}{x+1}+\frac{2}{x^{2}+1}+\frac{4}{x^{4}+1}+\ldots+\frac{2^{n}}{x^{2^{n}}+1} \text { is }
\]
(a) \( 2^{n} \frac{1}{1-x^{n+1}}-\frac{1}{(1-x)} \)
(b) \( 2^{n} \frac{1}{1-x^{2 n+1}} \)
(c) \( 2^{n+1} \frac{1}{1-x^{2 n+1}}-\frac{1}{(1-x)} \)
(d) \( 2^{n+1} \frac{1}{1-x^{2 n+1}} \)
\( \mathrm{P} \)
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