Q56 / 120 · VITEEE 2013 · Mathematics
The sum of the series
$\sum_{r=0}^{n}(-1)^{r n} C_{r}\left(\frac{1}{2^{r}}+\frac{3^{r}}{2^{2 r}}+\frac{7^{r}}{2^{3 r}}+\frac{15^{r}}{2^{4 r}}+\ldots m\right.$ terms $)$ is
- Correct answerA$\frac{2^{m n}-1}{2^{m n}\left(2^{n}-1\right)}$
- B$\frac{2^{m n}-1}{2^{n}-1}$
- C$\frac{2^{m n}+1}{2^{n}+1}$
- DNone of these
Solution
$\quad \sum_{\mathrm{r}=0}^{\mathrm{n}}(-1)^{\mathrm{r}} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$
$\left(\frac{1}{2^{\mathrm{r}}}+\frac{3^{\mathrm{r}}}{2^{2 \mathrm{r}}}+\frac{7^{\mathrm{r}}}{2^{3 \mathrm{r}}}+\ldots \text { upto } \mathrm{m} \text { terms }\right)$ $=\sum_{\mathrm{r}=0}^{\mathrm{n}}(-1)^{\mathrm{r}} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \frac{1}{2^{\mathrm{r}}}+\sum_{\mathrm{r}=0}^{\mathrm{n}}(-1)^{\mathrm{r}} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \cdot \frac{3^{\mathrm{r}}}{2^{2 \mathrm{r}}}$ $\quad+\sum_{\mathrm{r}=0}^{\mathrm{n}}(-1)^{\mathrm{r}} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \cdot \frac{7^{\mathrm{r}}}{2^{3 \mathrm{r}}}+\ldots$ $=\left(1-\frac{1}{2}\right)^{\mathrm{n}}+\left(1-\frac{3}{4}\right)^{\mathrm{n}}+\left(1-\frac{7}{8}\right)^{\mathrm{n}}+$ $\quad$
...upto $\mathrm{m}$ terms
$=\frac{1}{2^{\mathrm{n}}}+\frac{1}{4^{\mathrm{n}}}+\frac{1}{8^{\mathrm{n}}}+\ldots$ upto $\mathrm{m}$ terms
$=\frac{\frac{1}{2^{\mathrm{n}}}\left(1-\frac{1}{2^{\mathrm{m} \mathrm{n}}}\right)}{\left(1-\frac{1}{2^{\mathrm{n}}}\right)}=\frac{2^{\mathrm{mn}}-1}{2^{\mathrm{mn}}\left(2^{\mathrm{n}}-1\right)}$
Hence, the correct answer is option: 'A'.