$\sum _{r=0}^n{\left(-1\right)}^r {}^n{C}_r\left\{\frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+\frac{{15}^r}{2^{4r}}+\mathrm{......}\text{upto\hspace{0.17em}m\hspace{0.17em}\hspace{0.33em}terms}\right\}$

   $\because {(1-x)}^n=\sum _{r=0}^n{{(-1)}^r}^n{C}_r{x}^r$          .....(i)
Let   $P=\sum _{r=0}^n{(-1)}^r{}^n{C}_r\{{\left(\frac{1}{2}\right)}^r+{\left(\frac{3}{4}\right)}^r+{\left(\frac{7}{8}\right)}^r+{\left(\frac{15}{16}\right)}^r+\mathrm{.....}uptomterms\}$

$\begin{array}{l}=\sum _{r=0}^n{{(-1)}^r}^n{C}_r.{\left(\frac{1}{2}\right)}^r+\sum _{r=0}^n{{(-1)}^r}^n{C}_r.{\left(\frac{3}{4}\right)}^r\\ +\sum _{r=0}^n{{(-1)}^r}^n{C}_r.{\left(\frac{7}{8}\right)}^r+\sum _{r=0}^n{{(-1)}^r}^n{C}_r.{\left(\frac{15}{16}\right)}^r+\mathrm{.....}uptomterms\end{array}$

$={(1-\frac{1}{2})}^n+{(1-\frac{3}{4})}^n+{(1-\frac{7}{8})}^n+{(1-\frac{15}{16})}^n+\mathrm{.....}uptomterms$

$={\left(\frac{1}{2}\right)}^n+{\left(\frac{1}{2}\right)}^{2n}+{\left(\frac{1}{2}\right)}^{3n}+{\left(\frac{1}{2}\right)}^{4n}+\mathrm{.....}uptomterms$

$\frac{={\left(\frac{1}{2}\right)}^n[1-{\left\{{\left(\frac{1}{2}\right)}^m\right\}}^m]}{1-{\left(\frac{1}{2}\right)}^n}$

$=\frac{(2^{mn}-1)}{2^{mn}(2^n-1)}$