# Find the coefficient of xy^8 in the expansion of (x-y^2)^5?

May 17, 2017

Coefficient of $x {y}^{8}$ is $5$

#### Explanation:

Binomial expansion of ${\left(a - b\right)}^{n}$ is given by

${C}_{0}^{n} {a}^{n} {\left(- b\right)}^{0} + {C}_{1}^{n} {a}^{n - 1} {\left(- b\right)}^{1} + {C}_{2}^{n} {a}^{n - 2} {\left(- b\right)}^{2} + {C}_{3}^{n} {a}^{n - 3} {\left(- b\right)}^{3} + \ldots \ldots + {C}_{r}^{n} {a}^{n - r} {\left(- b\right)}^{r} + \ldots \ldots \ldots . + {C}_{n}^{n} {a}^{0} {\left(- b\right)}^{n}$, where C_r^n=(n!)/((n-r)!r!).

Hence ${\left(r + 1\right)}^{t h}$ term is ${C}_{r}^{n} {a}^{n - r} {\left(- b\right)}^{r}$

As we are given ${\left(x - {y}^{2}\right)}^{5}$, its ${\left(r + 1\right)}^{t h}$ term is

${C}_{r}^{5} {x}^{5 - r} {\left(- {y}^{2}\right)}^{r}$

As we are seeking coefficient of $x {y}^{8}$, it means $r = 4$

and corresponding term is ${C}_{4}^{5} {x}^{1} {\left(- {y}^{2}\right)}^{4}$

= $5 x {y}^{8}$ and coefficient of $x {y}^{8}$ is $5$

May 17, 2017

The coefficient is 5.

#### Explanation:

The question is asking what is the coefficient associated with the term $x {y}^{8}$ in the expansion of ${\left(x - {y}^{2}\right)}^{5}$. To make it easy on myself I used Pascal's triangle to help me in the expansion. The triangle has many interesting properties, but for our purposes we need to realise that each row gives the coefficients for the expansion of the general expression ${\left(a + b\right)}^{n}$

${\left(a + b\right)}^{0}$ 1
${\left(a + b\right)}^{1}$ 1 1
${\left(a + b\right)}^{2}$ 1 2 1
${\left(a + b\right)}^{3}$ 1 3 3 1
${\left(a + b\right)}^{4}$ 1 4 6 4 1
${\left(a + b\right)}^{5}$ 1 5 10 10 5 1
... and so on
So in our case if you let $a = x$ and $b = {y}^{2}$
${\left(x - {y}^{2}\right)}^{5} = {x}^{5} - 5 {x}^{4} {y}^{2} + 10 {x}^{3} {y}^{4} - 10 {x}^{2} {y}^{6} + 5 x {y}^{8} - {y}^{10}$