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Found in: Page 421

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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

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

See the step by step solution

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.

${\mathbf{\left(}}{\mathbit{x}}{\mathbf{+}}{\mathbit{y}}{\mathbf{\right)}}^{\mathbf{n}}{\mathbf{=}}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{0}}^{\mathbf{n}}{\mathbf{ }}\left(\begin{array}{l}n\\ j\end{array}\right){{\mathbit{x}}}^{\mathbf{n}\mathbf{-}\mathbf{j}}{{\mathbit{y}}}^{{\mathbf{j}}}$

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

$\begin{array}{r}n=200\\ j=99\end{array}$

Step 2: Find the corresponding term

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

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