Binomial Theorem 1 Question 32
34.
Find the sum of the series
$ \sum_{r=0}^{n}(-1)^{r}{ }^{n} C_{r}[ \frac{1}{2^{r}}+\frac{3^{r}}{2^{2 r}}+\frac{7^{r}}{2^{3 r}}+\frac{15^{r}}{2^{4 r}} \ldots \text { upto } m \text { terms }] $
$(1985,5 \mathrm{M})$
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Answer:
Correct Answer: 30. $(\frac{2^{m n}-1}{2^{m n}\left(2^{n}-1\right)})$
Solution:
- $\sum_{r=0}^{n}(-1)^{r}{ }^{n} C_{r}[ \frac{1}{2^{r}}+\frac{3^{r}}{2^{2 r}}+\frac{7^{r}}{2^{3 r}}+\frac{15^{r}}{2^{4 r}}+\ldots$ upto $m$ terms]
$ \begin{aligned} & =\sum_{r=0}^{n}(-1)^{r n} C_{r} (\frac{1}{2})^{r}+\sum_{r=0}^{n}(-1)^{r n} C_{r} (\frac{3}{4})^{r}+ \\ & \sum_{r=0}^{n}(-1)^{r}{ }^{n} C_{r} (\frac{7}{8})^{r}+\ldots \text { upto } m \text { terms } \end{aligned} $
$=(1-\frac{1}{2})^{n}+(1-\frac{3}{4})^{n}+(1-\frac{7}{8})^{n}+\ldots$ upto $m$ terms
[using $\sum_{r=0}^{n}(-1)^{r}{ }^{n} C_{r} x^{r}=(1-x)^{n}$]
$=(\frac{1}{2})^{n}+(\frac{1}{4})^{n}+(\frac{1}{8})^{n}+\ldots$ upto $m$ terms
$=(\frac{1}{2})^{n}[{\frac{1-({\frac{1}{2^{n}}})^{m}}{1-\frac{1}{2^{n}}}}]=\frac{2^{m n}-1}{2^{m n}\left(2^{n}-1\right)}$
