SATHEE: Sequences and Series 4 Question 2

Sequences and Series 4 Question 2

2.

Let $S_{n} = 1 + q + q^{2} + \dots + q^{n}$ and $T_{n} = 1 + (\frac{q + 1}{2}) + (\frac{q + 1}{2})^{2} + \dots + (\frac{q + 1}{2})^{n}$ , where $q$ is a real number and $q \neq 1$ . If

$^{101} C_{1} +^{101} C_{2} \cdot S_{1} + \dots +^{101} C_{101} \cdot S_{100} = α T_{100}$ , then $α$ is equal to

(a) $2^{100}$

(b) 202

(d) $2^{99}$

(2019 Main, 11 Jan II)

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Answer:

Correct Answer: 2. (a)

Solution:

(a) We have, $S_{n} = 1 + q + q^{2} + \dots + q^{n}$ and

$T_{n} = 1 + (\frac{q + 1}{2}) + (\frac{q + 1}{2})^{2} + \dots + (\frac{q + 1}{2})^{n}$

Also, we have

$\begin{aligned} ^{101} C_{1} +^{101} C_{2} S_{1} +^{101} C_{3} S_{2} + \dots +^{101} C_{101} S_{100} = α T_{100} \\ \Rightarrow^{101} C_{1} +^{101} C_{2} (1 + q) +^{101} C_{3} (1 + q + q^{2}) \\ + \dots +^{101} C_{101} (1 + q + q^{2} + \dots + q^{100}) \\ = α \cdot T_{100} \\ \Rightarrow^{101} C_{1} +^{101} C_{2} \frac{(1 - q^{2})}{1 - q} +^{101} C_{3} \frac{1 - q^{3}}{1 - q} \\ +^{101} C_{4} \frac{1 - q^{4}}{1 - q} + \dots +^{101} C_{101} \frac{1 - q^{101}}{1 - q} \\ = α \cdot T_{100} [∵ for a GP, S_{n} = a \frac{1 - r^{n}}{1 - r}, r \neq 1] \\ \Rightarrow \frac{1}{1 - q} [^{101} C_{1} +^{101} C_{2} + \dots +^{101} C_{101} \\ -^{101} C_{1} q +^{101} C_{2} q^{2} + \dots +^{101} C_{101} q^{101} = α \cdot T_{100} \\ \Rightarrow \frac{1}{(1 - q)} [(2^{101} - 1) - ((1 + q)^{101} - 1)] = α T_{100} \end{aligned}$

$\begin{aligned} [∵ q \neq 1 \Rightarrow q + 1 \neq 2 \Rightarrow \frac{q + 1}{2} \neq 1] \\ = \frac{α [2^{101} - (q + 1)^{101}]}{(1 - q) \cdot 2^{100}} \Rightarrow α = 2^{100} \end{aligned}$

Sequences and Series 4 Question 2

2.

Answer:

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Sequences and Series 4 Question 2

2.

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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