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

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Answer 1 · 2014-10-02T08:31:58

By taking the derivative of a geometric power series,

#f(x)=sum_{n=1}^infty nx^{n-1}#.

Let us look at some details.

Since

#1/{1-x}=sum_{n=0}^infty x^n#,

by taking the derivative,

#1/{(1-x)^2}=sum_{n=0}^infty nx^{n-1}=sum_{n=1}^infty nx^{n-1}#.

I hope that this was helpful.