Question

The sum of the series $ \displaystyle\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)

$\displaystyle\sum_{r=0}^{n}(-1)^{r \cdot n} C_{r}$
$\left(\frac{1}{2^{r}}+\frac{3^{r}}{2^{2 r}}+\frac{7^{r}}{2^{3 r}}+\ldots\right.$ upto $m$ terms $)$
$=\displaystyle\sum_{r=0}^{n}(-1)^{r} \cdot{ }^{n} C_{r} \frac{1}{2^{r}}+\displaystyle\sum_{r=0}^{n}(-1)^{r} \cdot{ }^{n} C_{r} \cdot \frac{3^{r}}{2^{2 r}}$
$+\displaystyle\sum_{r=0}^{n}(-1)^{r} \cdot{ }^{n} C_{r} \cdot \frac{7^{r}}{2^{3 r}}+\ldots$
$=\left(1-\frac{1}{2}\right)^{n}+\left(1-\frac{3}{4}\right)^{n}+\left(1-\frac{7}{8}\right)^{n}+$ upto $m$ terms
$=\frac{1}{2^{n}}+\frac{1}{4^{n}}+\frac{1}{8^{n}}+\ldots$ upto $m$ terms
$=\frac{\frac{1}{2^{n}}\left(1-\frac{1}{2^{m n}}\right)}{\left(1-\frac{1}{2^{n}}\right)}=\frac{2^{m n}-1}{2^{m n}\left(2^{n}-1\right)}$