# How do you find the coefficient of a of the term ax^2y^8 in the expansion of the binomial (4x-y)^10?

Mar 5, 2017

Coefficient $a$ in the term $a {x}^{8} {y}^{2}$ is $720$.

#### Explanation:

The terms of a binomial expansion of ${\left(x + a\right)}^{n}$ are given by the series (with $\left(n + 1\right)$ terms

${\Sigma}_{r = 0}^{r = n} \left(\begin{matrix}n \\ r\end{matrix}\right) {x}^{n - r} {a}^{r}$, where ((n),(r))=(n!)/((n-r)!r!)

Hence expansion of ${\left(4 x - y\right)}^{10}$ is given by

${\Sigma}_{r = 0}^{r = 10} \left(\begin{matrix}10 \\ r\end{matrix}\right) {\left(4 x\right)}^{10 - r} {\left(- y\right)}^{r}$

As we are coefficient of $a$ of the term $a {x}^{2} {y}^{8}$

$10 - r = 2$ or $r = 8$ and it will occur in the ${\left(8 + 1\right)}^{t h}$ term of the sequence and this term will be

$\left(\begin{matrix}10 \\ 8\end{matrix}\right) {\left(4 x\right)}^{10 - 8} {\left(- y\right)}^{8}$

= (10!)/((10-8)!8!)(4x)^2(-y)^8#

= $\frac{10 \times 9}{1 \times 2} 16 {x}^{2} {y}^{8}$

= $720 {x}^{2} {y}^{8}$

Hence, coefficient $a$ in the term $a {x}^{8} {y}^{2}$ is $720$.