Question

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

Answer 1

The expansion is `x^4/(1-3x)^3=x^4+9x^5+54x^6+270x^7+cdots`, and is valid for `|x|<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/(1-3x)^3=x^4 * (1+(-3x))^(-3)` and use the expansion above with `z=-3x` 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(1+9x+6*9x^2-10*(-27)x^3+cdots)`

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