From David J. Griffiths "Introduction to Electrodynamics 2. ed":

"The third term" is further up the page.We're interested in the regime $\displaystyle r\gg s$ (...), so the third term is necligible, and the binomial expansion yields

$\displaystyle (...)\cong\frac{1}{r}\left(1\mp\frac{s}{r}\cos\the ta\right)^{-1/2}\cong\frac{1}{r}\left(1\pm\frac{s}{2r}\cos\theta \right)$

What is really going on here? I've been trying to understand it, but I can't figure it out. How does he get $\displaystyle \frac{1}{r}\left(1\pm\frac{s}{2r}\cos\theta\right)$?

Edit/context:

This is a step in deriving the formula

$\displaystyle V(P)\cong\frac{1}{4\pi\epsilon_{0}}\frac{qs\cos\th eta}{r^{2}}$

which in turn leads to

$\displaystyle V_{dip}(P)=\frac{1}{4\pi\epsilon_{0}}\frac{\mathbf {P}\cdot\mathbf{\hat{r}}}{r^{2}}$