Short Answer

$x^{101} y^{99}$ What is the coefficient of in the expansion of $(2 x - 3 y)^{200} ?$

The coefficient $x^{101} y^{99} is then - \frac{200!}{99! (200 - 99)!} 2^{101} 3^{99} \approx - 39 \times 10^{135}$

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Step by Step Solution

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TABLE OF CONTENTS

Step 1: Use Binomial theorem

Binomial theorem: binomial theorem, statement that for any positive integer n , the nth power of the sum of two numbers x and y may be expressed as the sum of terms of the form.

$(x + y)^{n} = \sum_{j = 0}^{n} (\begin{array}{l} n \\ j \end{array}) x^{n - j} y^{j}$

The term is $x^{101} y^{99} in (2 x - 3 y)^{200} = (2 x + (- 3 y))^{200}$

$\begin{aligned} n = 200 \\ j = 99 \end{aligned}$

Step 2: Find the corresponding term

$\begin{aligned} (\begin{array}{c} n \\ j \end{array}) (2 x)^{n - j} (- 3 y)^{j} = (\begin{array}{c} 200 \\ 99 \end{array}) (2 x)^{200 - 99} (- 3 y)^{99} \\ = - \frac{200!}{99! (200 - 99)!} (2 x)^{101} (3 y)^{99} \\ = - \frac{200!}{99! (200 - 99)!} 2^{101} x^{101} 3^{99} y^{99} \\ = - \frac{200!}{99! (200 - 99)!} 2^{101} 3^{99} x^{101} y^{99} \\ (\begin{array}{l} n \\ j \end{array}) (2 x)^{n - j} (- 3 y)^{j} \approx - 39 \times 10^{135} x^{101} y^{99} \end{aligned}$

Thus, the coefficient $x^{101} y^{99} is then - \frac{200!}{99! (200 - 99)!} 2^{101} 3^{99} \approx - 39 \times 10^{135}$

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Discrete Mathematics and its Applications

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Short Answer

$x^{101} y^{99}$ What is the coefficient of in the expansion of $(2 x - 3 y)^{200} ?$

Step by Step Solution

Step 1: Use Binomial theorem

Step 2: Find the corresponding term

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Discrete Mathematics and its Applications

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x101y99What is the coefficient of in the expansion of (2x−3y)200?

Step by Step Solution

Step 1: Use Binomial theorem

Step 2: Find the corresponding term

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$x^{101} y^{99}$ What is the coefficient of in the expansion of $(2 x - 3 y)^{200} ?$