# How do you use the binomial series to expand x^4/(1-3x)^3?

Sep 17, 2017

The expansion is ${x}^{4} / {\left(1 - 3 x\right)}^{3} = {x}^{4} + 9 {x}^{5} + 54 {x}^{6} + 270 {x}^{7} + \cdots$, and is valid for $| x | < \frac{1}{3}$.

#### Explanation:

The general binomial series expansion can be written as (1+z)^p=1+pz+(p(p-1))/(2!)z^2+(p(p-1)(p-2))/(3!)z^3+cdots for $| z | < 1$ (though the expansion is finite and works for all $z$ if $p$ is a non-negative integer).

For the given expression, we can write
${x}^{4} / {\left(1 - 3 x\right)}^{3} = {x}^{4} \cdot {\left(1 + \left(- 3 x\right)\right)}^{- 3}$ and use the expansion above with $z = - 3 x$ and $p = - 3$. This gives:

x^4 * (1+(-3x))^(-3)=x^4(1-3(-3x)+((-3)*(-4))/(2!)(-3x)^2+

((-3) * (-4) * (-5))/(3!)(-3x)^3+cdots).

This simplifies to

${x}^{4} \left(1 + 9 x + 6 \cdot 9 {x}^{2} - 10 \cdot \left(- 27\right) {x}^{3} + \cdots\right)$

$= {x}^{4} + 9 {x}^{5} + 54 {x}^{6} + 270 {x}^{7} + \cdots$, and is valid for $| x | < \frac{1}{3}$ (which is equivalent to $| z | = | - 3 x | < 1$).