Question

The sum of the series $\sum _{r=0}^n{\left(-1\right)}^r{}^n{C}_r\left(\frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+\frac{15^r}{2^{4r}}+...m\text{terms}\right)$ is

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

Answer 1

Answer: (A)

$\sum _{r=0}^n(-1{)}^{r\cdot n}{C}_r$
$(\frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+\dots$ upto $m$ terms $)$
$=\sum _{r=0}^n(-1{)}^r\cdot {}^n{C}_r\frac{1}{2^r}+\sum _{r=0}^n(-1{)}^r\cdot {}^n{C}_r\cdot \frac{3^r}{2^{2r}}$
$+\sum _{r=0}^n(-1{)}^r\cdot {}^n{C}_r\cdot \frac{7^r}{2^{3r}}+\dots$
$={\left(1-\frac{1}{2}\right)}^n+{\left(1-\frac{3}{4}\right)}^n+{\left(1-\frac{7}{8}\right)}^n+$ upto $m$ terms
$=\frac{1}{2^n}+\frac{1}{4^n}+\frac{1}{8^n}+\dots$ upto $m$ terms
$=\frac{\frac{1}{2^n}\left(1-\frac{1}{2^{mn}}\right)}{\left(1-\frac{1}{2^n}\right)}=\frac{2^{mn}-1}{2^{mn}\left(2^n-1\right)}$