Question

The sum of the series $\sum _{r=0}^n{(-1)}^r{}^n{C}_r[\frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+\frac{{15}^r}{2^{4r}}+\cdots uptopterms]$

• A. $\frac{2^{pn}-1}{2^{np}(2^n+1)}$
• B. $\frac{2^{pn}-1}{2^{np}(2^n-1)}$
• C. $\frac{2^{2pn}-1}{2^{np}(2^n-1)}$
• D. None of these

Answer 1

The correct option is B $\frac{2^{pn}-1}{2^{np}(2^n-1)}$

$\sum _{r=0}^n{(-1)}^r{}^n{C}_r[\frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+\frac{{15}^r}{2^{4r}}+...]$ upto $m$ terms
$=\sum _{r=0}^n{(-1)}^r{\left(\frac{1}{2}\right)}^r+\sum _{r=0}^n{(-1)}^r{}^n{C}_r{\left(\frac{3}{4}\right)}^r+\sum _{r=0}^n{(-1)}^r{{}^{n}C}_r{\left(\frac{7}{8}\right)}^r+...$ upto $m$ terms
$={(1-\frac{1}{2})}^n+{(1-\frac{3}{4})}^n+{(1-\frac{7}{8})}^n+...$ upto $m$ terms
using $\{{\sum }_{r=0}^n{(-1)}^r{}^n{C}_r{x}^r={(1-x)}^n\}$
$={\left(\frac{1}{2}\right)}^n+{\left(\frac{1}{4}\right)}^n+{\left(\frac{1}{8}\right)}^n+...$ upto $m$ terms
$={\left(\frac{1}{2}\right)}^n\left[\frac{1-{\left(\frac{1}{2^n}\right)}^m}{1-\frac{1}{2^n}}\right]=\frac{2^{mm}-1}{2^{mm}(2^n-1)}$