SATHEE: Binomial Theorem 2 Question 1

Binomial Theorem 2 Question 1

1.

Let $(x + 10)^{50} + (x - 10)^{50} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{50} x^{50}$ , for all $x \in R$ ; then $\frac{a_{2}}{a_{0}}$ is equal to

(2019 Main, 11 Jan II)

(a) $12.25$

(b) $12.50$

(d) $12.75$

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Answer:

Correct Answer: 1. (a)

Solution:

We have,

$(x + 10)^{50} + (x - 10)^{50} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{50} x^{50}$

$∴ a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{50} x^{50}$

$= [^{50} C_{0} x^{50} +^{50} C_{1} x^{49} 10 +^{50} C_{2} x^{48} \cdot 10^{2} + \dots +^{50} C_{50} 10^{50}]$ +

$[^{50} C_{0} x^{50} -^{50} C_{1} x^{49} 10 +^{50} C_{2} x^{48} 10^{2} - \dots +^{50} C_{50} 10^{50}]$

$= 2 [^{50} C_{0} x^{50} +^{50} C_{2} x^{48} \cdot 10^{2} +^{50} C_{4} x^{46} \cdot 10^{4}$ $+ \dots +^{50} C_{50} \cdot 10^{50}]$

By comparing coefficients, we get

$a_{2} = 2^{50} C_{48} (10)^{48}; a_{0} = 2^{50} C_{50} (10)^{50} = 2 (10)^{50}$

$∴ \frac{a_{2}}{a_{0}} = \frac{2 (^{50} C_{2}) (10)^{48}}{2 (10)^{50}} = 2 \frac{50 \cdot 49}{1 \cdot 2} \frac{(10)^{48}}{2 \cdot (10)^{50}}$

$[∴^{50} C_{48} =^{50} C_{2}]$

$= \frac{50 \times 49}{2 \cdot (10 \times 10)} = \frac{5 \times 49}{20} = \frac{245}{20} = 12.25$

Binomial Theorem 2 Question 1

1.

Answer:

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Binomial Theorem 2 Question 1

1.

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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