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The sum to $\mathrm{n}$ terms of the series
$\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots \ldots \ldots \ldots$ is
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$\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots \ldots \ldots \ldots$ is
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The correct answer is:
$n-1+2^{-n}$
$\begin{aligned} \frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots \\=&\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+\left(1-\frac{1}{16}\right)+\ldots . \\=& \mathrm{n}-\frac{\frac{1}{2}\left\{1-\frac{1}{2^{\mathrm{n}}}\right\}}{1-\frac{1}{2}}=\mathrm{n}-1+2^{-\mathrm{n}} \end{aligned}$
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