Question

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

• A. $\frac{2^n-1}{2^{mn}(2^n-1)}$
• B. $\frac{2^m-1}{2^n(2^n-1)}$
• C. $\frac{2^{mn}-1}{2^{mn}(2^n-1)}$
• D. $\frac{2^{mn}-1}{2^{mn}(2^m-1)}$

Answer 1

The correct option is C

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

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