Series

C.$n-1+2^{-n}$

Given $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots$
$$
\begin{aligned}
& =\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+\left(1-\frac{1}{16}\right)+\ldots \\
& =\mathrm{n}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\ldots \ldots \ldots+\frac{1}{2^n}\right)
\end{aligned}
$$
$=\mathrm{n}-\frac{\frac{1}{2}\left(1-2^{-\mathrm{n}}\right)}{1-\frac{1}{2}} \quad\left[\right.$ Sum of $\mathrm{GP}=\frac{\mathrm{a}\left(1-\mathrm{r}^{\mathrm{n}}\right)}{1-\mathrm{r}}$ when $\left.\mathrm{r}<1\right]$
$$
=n-1+2^{-n}
$$