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Taylor expansion of $$\frac{1}{(x^2-2x+17)^2}$$ at point $c=1$ ?

My attempt: $y = x -c = x-1$ so $x = y +1 $. When we substitute this we get the expression $\frac{1}{(x^2+16)^2}$. Now, I have the formula $(1+x)^\alpha = \sum_{n=0}^{\infty}{\binom{\alpha}{n}x^n},\lvert x \rvert < 1$

I was wondering if I could write my expression as $4((\frac{x}{4})^2+1)^{-2}$ and apply the formula, or should I take a different approach? Am I even allowed to do that?

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Let $z=(x-1)$ \begin{eqnarray*} \frac{1}{(x^2-2x+17)^2}=\frac{1}{(z^2+16)^2} =\frac{1}{16^2} \left(1+\frac{z^2}{16} \right)^{-2} \end{eqnarray*} $\left(1+\frac{z^2}{16} \right)^{-2}$ is now ripe to apply the binomial expansion \begin{eqnarray*} \left(1+\frac{z^2}{16} \right)^{-2}= 1+(-2) \frac{z^2}{16}+\frac{(-2)(-2-1)}{2!}\left(\frac{z^2}{16} \right)^{-2}+\cdots \end{eqnarray*} and so we have \begin{eqnarray*} \frac{1}{(x^2-2x+17)^2}=\frac{1}{16^2} \left(1-\frac{(x-1)^2}{8} +\frac{3(x-1)^4}{16^2} +\cdots \right) \end{eqnarray*} and this series is valid provided $ \mid x-1 \mid <4$.

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