How does one expand $\sqrt{r^2+a^2}-r$ to $\frac{a^2}{2r}-\frac{a^4}{8r^3}+\frac{a^6}{16r^5}$? The problem and solution are attached as photos below. 
I understand all parts of the solution except for the line above "By inspection..." where $V(r,\theta=0)$ is expanded. I tried a taylor expansion, taylor expansion + binomial, etc, to no avail, but maybe I don't know what I'm doing. Help is much appreciated!
Remember we assume r>~a in part b. The solution comes from a trusted source. 



Taylor expansion is indeed the key here, but you have to be mindful of what variable you are expanding! Here, depending on how far $(r)$ you are away from the disk, the smaller the disk is to you and less its extended structure matters so $a$ is 'small' for your purposes. We then should Taylor expand around $a=0$.
You don't want to Taylor expand around $r=0$ because that would be assuming that $r$ is small but it's given in the problem that $r>a$. 
So we say
\begin{equation}
f(a) = \sqrt{r^2+a^2} = f(0) + f'(0)~a + \frac{1}{2}f''(0)~a^2+ \frac{1}{3!}f'''(0)~a^3 + \frac{1}{4!}f^{(4)}(0)~a^4 + \theta(a^5)
\end{equation}
Now the derivatives of $\sqrt{r^2+a^2}$ are
\begin{align}
\frac{\partial}{\partial a}\sqrt{a^2+r^2} = \frac{a}{\sqrt{r^2+a^2}}~,\\
\frac{\partial^2}{\partial a^2}\sqrt{a^2+r^2} = \frac{1}{\sqrt{r^2+a^2}}-\frac{a^2}{(r^2+a^2)^{3/2}}~,\\
\frac{\partial^3}{\partial a^3}\sqrt{a^2+r^2} = - \frac{3a r^2}{(r^2+a^2)^{5/2}}~,\\
\frac{\partial^4}{\partial a^4}\sqrt{a^2+r^4} = \frac{12 a^2 r^2}{(r^2+a^2)^{7/2}} - \frac{3r^4}{(r^2+a^2)^{7/2}}\\
...
\end{align}
so evaluating these at $a=0$ (only the 2nd and 4th derivatives are non-zero at $a=0$) and plugging into our Taylor expansion we get
$$
\sqrt{r^2+a^2} = \frac{a^2}{2r} - \frac{a^4}{8r^3} + \theta(a^5)~~.
$$
I'll leave it as an exercise to you to calculate the 5th and 6th derivatives with respect to $a$ to see if you can match the expression from your solution.
