Why $\frac{1}{w-z}(\frac{1}{(\xi-w)^{\ell+1}}-\frac{1}{(\xi-z)^{\ell+1}})= \frac{\sum_{k=0}^{\ell}(\xi-z)^{\ell-k}(\xi-w)^k}{(\xi-w)^{\ell+1}(\xi-z)^{\ell+1}}$ for a natural number $\ell$? It comes from some calculation in complex analysis.
In my opinion, $\frac{1}{(\xi-w)^{\ell+1}}-\frac{1}{(\xi-z)^{\ell+1}}=\frac{(\xi-z)^{\ell+1}-(\xi-w)^{\ell+1}}{(\xi-w)^{\ell+1}(\xi-z)^{\ell+1}}$ and I try to use $(x+y)^n=\sum_{i=0}^n\binom{n}{i}x^iy^{n-i}$. But it seems lots of computations are involved. I am wondering if there is a fast way to figure it out?
Essentially, use the geometric sum formula.
Looking at the numerator: $\begin{array}{rcl} (\xi-z)^{l+1}-(\xi-w)^{l+1}&=&(\xi-z)^{l+1}\left(1-\left(\dfrac{\xi-w}{\xi-z}\right)^{l+1}\right)\\ &=&(\xi-z)^{l+1}\displaystyle\sum_{k=0}^l\left(\dfrac{\xi-w}{\xi-z}\right)^{k}\left(1-\dfrac{\xi-w}{\xi-z} \right)\\ &=&(\xi-z)^{l+1}\displaystyle\sum_{k=0}^l\left(\dfrac{\xi-w}{\xi-z}\right)^{k}\left(\dfrac{w-z}{\xi-z} \right)\\ &=&\displaystyle\sum_{k=0}^l(\xi-z)^{l-k}\left(\xi-w\right)^{k}(w-z) \end{array}.$
The result follows.