# How do you find the Maclaurin series of f(x)=(1-x)^-2 ?

Oct 2, 2014

By taking the derivative of a geometric power series,

$f \left(x\right) = {\sum}_{n = 1}^{\infty} n {x}^{n - 1}$.

Let us look at some details.

Since

$\frac{1}{1 - x} = {\sum}_{n = 0}^{\infty} {x}^{n}$,

by taking the derivative,

$\frac{1}{{\left(1 - x\right)}^{2}} = {\sum}_{n = 0}^{\infty} n {x}^{n - 1} = {\sum}_{n = 1}^{\infty} n {x}^{n - 1}$.

I hope that this was helpful.