First slide

Binomial theorem for positive integral Index

Question

The sum of the series $\sum_{r = 0}^{n} (- 1)^{r^{n}} C_{r} (\frac{1}{2^{r}} + \frac{3^{r}}{2^{2 r}} + \frac{7^{r}}{2^{3 r}} + \frac{15^{r}}{2^{4 r}} + \dots + m terms) is$

Moderate

A

$\frac{2^{mn} - 1}{2^{mn} (2^{n} - 1)}$
B

$\frac{2^{mn} - 1}{2^{n} - 1}$
C

$\frac{2^{mn} + 1}{2^{n} + 1}$
D

None of these

Solution

$\sum_{n}^{n} (- 1)^{rn} C_{r} \{\frac{1}{2^{r}} + \frac{3^{r}}{2^{2 r}} + \frac{7^{r}}{2^{3 r}} + \dots upto m terms\}$
$= \sum_{r = 0}^{n} (- 1)^{r n} C_{r} \cdot \frac{1}{2^{r}} + \sum_{r = 0}^{n} (- 1)^{r} \cdot^{n} C_{r} \frac{3^{r}}{2^{2 r}}$ $+ \sum_{r = 0}^{n} (- 1)^{rn} C_{r} \frac{7^{r}}{2^{3 r}} + \dots$
$= {(1 - \frac{1}{2})}^{n} + {(1 - \frac{3}{4})}^{n} + {(1 - \frac{7}{8})}^{n} + \dots$ up to M terms
$= \frac{1}{2^{n}} + \frac{1}{2^{2 n}} + \frac{1}{2^{3 n}} \dots upto m terms$
$= \frac{\frac{1}{2^{n}} \{1 - {(\frac{1}{2^{n}})}^{m}\}}{(1 - \frac{1}{2^{n}})} = \frac{2^{mn} - 1}{2^{mn} (2^{n} - 1)}$

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