SATHEE: Binomial Theorem 1 Question 32

Binomial Theorem 1 Question 32

34.

Find the sum of the series

$\sum_{r = 0}^{n} (- 1)^{r}^{n} C_{r} [\frac{1}{2^{r}} + \frac{3^{r}}{2^{2 r}} + \frac{7^{r}}{2^{3 r}} + \frac{15^{r}}{2^{4 r}} \dots upto m terms]$

$(1985, 5 M)$

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Answer:

Correct Answer: 30. $(\frac{2^{m n} - 1}{2^{m n} (2^{n} - 1)})$

Solution:

$\sum_{r = 0}^{n} (- 1)^{r}^{n} C_{r} [\frac{1}{2^{r}} + \frac{3^{r}}{2^{2 r}} + \frac{7^{r}}{2^{3 r}} + \frac{15^{r}}{2^{4 r}} + \dots$ upto $m$ terms]

$\begin{aligned} = \sum_{r = 0}^{n} (- 1)^{r n} C_{r} (\frac{1}{2})^{r} + \sum_{r = 0}^{n} (- 1)^{r n} C_{r} (\frac{3}{4})^{r} + \\ \sum_{r = 0}^{n} (- 1)^{r}^{n} C_{r} (\frac{7}{8})^{r} + \dots upto m terms \end{aligned}$

$= (1 - \frac{1}{2})^{n} + (1 - \frac{3}{4})^{n} + (1 - \frac{7}{8})^{n} + \dots$ upto $m$ terms

[using $\sum_{r = 0}^{n} (- 1)^{r}^{n} C_{r} x^{r} = (1 - x)^{n}$ ]

$= (\frac{1}{2})^{n} + (\frac{1}{4})^{n} + (\frac{1}{8})^{n} + \dots$ upto $m$ terms

$= (\frac{1}{2})^{n} [\frac{1 - (\frac{1}{2^{n}})^{m}}{1 - \frac{1}{2^{n}}}] = \frac{2^{m n} - 1}{2^{m n} (2^{n} - 1)}$

Binomial Theorem 1 Question 32

34.

Answer:

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Binomial Theorem 1 Question 32

34.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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