SATHEE: Binomial Theorem 1 Question 33

Binomial Theorem 1 Question 33

35.

Given, $s_{n} = 1 + q + q^{2} + \dots + q^{n}$

Prove that $^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2}$ $+ \dots +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$

$(1984, 4$ M)

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Solution:

$^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + \dots +^{n + 1} C_{n + 1} s_{n}$

$= \sum_{r = 1}^{n + 1}^{n + 1} C_{r} s_{r - 1}$

where $s_{n} = 1 + q + q^{2} + \dots + q^{n} = \frac{1 - q^{n + 1}}{1 - q}$

$\begin{aligned} ∴ \sum_{r = 1}^{n + 1}^{n + 1} C_{r} \frac{1 - q^{r}}{1 - q} & = \frac{1}{1 - q} \sum_{r = 1}^{n + 1}^{n + 1} C_{r} - \sum_{r = 1}^{n + 1}^{n + 1} C_{r} q^{r} \\ = \frac{1}{1 - q} [(1 + 1)^{n + 1} - (1 + q)^{n + 1}] \\ = \frac{1}{1 - q} [2^{n + 1} - (1 + q)^{n + 1}] . \dots \dots . (i) \end{aligned}$

$= \frac{1 - (\frac{q + 1}{2})^{n + 1}}{1 - (\frac{q + 1}{2})} = \frac{2^{n + 1} - (q + 1)^{n + 1}}{2^{n} (1 - q)} \dots \dots . (i i)$

From Eqs. (i) and (ii),

$^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + \dots +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$

Binomial Theorem 1 Question 33

35.

Solution:

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Binomial Theorem 1 Question 33

35.

Solution:

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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