Binomial Theorem 1 Question 32
34. Find the sum of the series
$$ \sum _{r=0}^{n}(-1)^{r}{ }^{n} C _r \frac{1}{2^{r}}+\frac{3^{r}}{2^{2 r}}+\frac{7^{r}}{2^{3 r}}+\frac{15^{r}}{2^{4 r}} \ldots \text { upto } m \text { terms } $$
$(1985,5 M)$
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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
$=\frac{1}{2} \frac{n}{\frac{1-{\frac{1}{2^{n}}}^{m}}{1-\frac{1}{2^{n}}}}=\frac{2^{m n}-1}{2^{m n}\left(2^{n}-1\right)}$
