Question

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

Short answer

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

Step-by-step solution

Use Binomial theorem

Binomial theorem: binomial theorem, statement that for any positive integer n, the nth power of the sum of two numbers xand ymay be expressed as the sum ofterms of the form.

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

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

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

Find the corresponding term

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

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