Question

What is the coefficient of${\mathit{x}}^{\mathbf{8}}{\mathit{y}}^{\mathbf{9}}$in the expansion of$\mathbf{(}\mathbf{3}\mathit{x}\mathbf{+}\mathbf{2}\mathit{y}{\mathbf{)}}^{\mathbf{17}}$?

Short answer

$\text{The coefficient of}{x}^8{y}^9\text{is}81,662,929,920\text{.}$

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 of n + 1terms 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$x^8{y}^9\text{in}(3x+2y{)}^{17}$

$\begin{array}{r}n=17\\ j=9\end{array}$

Find the corresponding term

$\begin{array}{r}\left(\begin{array}{c}n\\ j\end{array}\right)\left(3x{)}^{n-j}\right(2y{)}^j=\left(\begin{array}{c}17\\ 9\end{array}\right)\left(3x{)}^{17-9}\right(2y{)}^9\\ =\frac{17!}{9!(17-9)!}\left(3x{)}^8\right(2y{)}^9\\ =\frac{17!}{9!8!}{3}^8{x}^8{2}^9{y}^9\\ =24310\cdot 3^8{2}^9{x}^8{y}^9\\ \left(\begin{array}{c}n\\ j\end{array}\right)\left(3x{)}^{n-j}\right(2y{)}^j=81,662,929,920x^8{y}^9\end{array}$

Thus, the coefficient of$x^8{y}^9$ is 81,662,929,920.