the taylor series of the function $f(x) = A/(x-B)^4$ using geometric series.

Question

I have to find the taylor series of the function $f(x) = A/(x-B)^4$ using geometric series.
If rewrote it to the general geometric series $\sum x^n=\frac{1}{1-x}$
$A\frac{1}{x-B}=A\frac{-\frac{1}{B}}{1-\frac{x}{B}}= A\cdot-\frac{1}{B}\sum(\frac{x}{B})^n$
There is one problem because now i dont have the power four yet. Is it possible to just set both sides to the power 4?
$A\left(\frac{1}{x-B}\right)^4=A\left(\frac{-\frac{1}{B}}{1-\frac{x}{B}}\right)^4= A\cdot(\frac{1}{B})^4\sum(\frac{x}{B})^{4n}$
This feels wrong so i wasnt sure.

Answer 1

One has $\sum_{n=0}^\infty x^n=\frac1{1-x}$. Differentiate three times: then we get $$\sum_{n=3}^\infty n(n-1)(n-2)x^n=\frac{6}{(1-x)^4}.$$ One gets a series for $1/(1-x)^4$ and one can massage it into a series for $1/(B-x)^4$.