Taylor Series of $ \frac{1}{x^2} $ about $ x = 1 $

Question

I am trying to solve the following problem:

$$ \text{Find taylor series of } \cfrac{1}{x^2} \text{ at } x = 1 $$

I tried to do the following:

$$ \begin{align} \cfrac{1}{(x-1)^2} &= \cfrac{1}{x^2-2x+1} \\ &= \cfrac{1}{1-x(2-x)} \end{align} $$

I know that I must obtain the form $ \cfrac{a}{1-(x-2)} $ but I just can't see how to transform the above equation. Could someone please advise me?

Answer 1

Observe \begin{align} \frac{-1}{(1+x)^2}=\frac{d}{dx}\frac{1}{1+x} = \sum^\infty_{n=1} (-1)^nnx^{n-1} \end{align} whenever $|x|<1$. Observe \begin{align} \frac{1}{x^2} = \frac{1}{(1+(x-1))^2} = \sum^\infty_{n=1}(-1)^{n+1} n(x-1)^{n-1}. \end{align}