Find first five non-zero terms of power series I have the function f(x) = $\dfrac{2x}{\left(x-5\right)^2}$, and I'm supposed to "find the first five non-zero terms of power series representation centered at x = 0."

Using $\frac{f^{(n)}(0)}{n!}$, the first non-zero terms I get are:

C1 = 2/25

C2 = 4/125

C3 = 6/625

C4 = 8/3125

C5 = 2/3125

However, WebWork marks these as incorrect. Any alternative methods or corrections to the work above are gladly appreciated.
Hint:

You don't have to compute the successive derivatives. Instead rewrite the fraction as
$$\frac{2x}{\left(x-5\right)^2}=\frac{2x}{25}\frac 1{\Bigl(1-\cfrac x5\Bigr)^2}=\frac{2x}5\left(\frac 1{1-\smash[b]{\cfrac x5}}\right)'$$
and use the power series expansion $\;\dfrac1{1-u}=1+u+u^2+\dotsm$
$\begin{array}\\
\dfrac{2x}{\left(x-5\right)^2}
&amp;=\dfrac{2x}{x^2\left(1-5/x\right)^2}\\
&amp;=\dfrac{2}{x}\sum_{n=0}^{\infty} (n+1)(5/x)^n\\
&amp;=\dfrac{2}{x}(1+2\dfrac{5}{x}+3(\dfrac{5}{x})^2+4(\dfrac{5}{x})^3+5(\dfrac{5}{x})^4+...)\\
&amp;=\dfrac{2}{x}(1+\dfrac{10}{x}+\dfrac{75}{x^2}+\dfrac{500}{x^3}+\dfrac{3125}{x^4}+...)\\
&amp;=\dfrac{2}{x}+\dfrac{20}{x^2}+\dfrac{150}{x^3}+\dfrac{1000}{x^4}+\dfrac{6250}{x^5}+...)\\
\end{array}
$