Multipole Expansion: Electrostatics Why in multipole expansion (or the terms therein) goes as mono-, di-, quadru-, octu-, or more specifically why are they in powers of 2? Why can't we have hexapole for instance?
 A: *

*When one solves Laplace's equation $$0~=~\nabla^2\Phi~=~\frac{1}{r^2}\frac{\partial}{\partial r}r^2 \frac{\partial \Phi}{\partial r} - \frac{1}{r^2}L^2\Phi $$in spherical coordinates in the bulk (away from a central region with sources), it separates in an angular problem on the $2$-sphere $S^2$ and a radial problem. 

*Since the rotation group $SO(3)$ acts on the $2$-sphere $S^2$, the angular solutions are representations of the Lie group $SO(3)$, namely linear combinations of the spherical harmonics. All finite-dimensional irreducible representations $V_{\ell}$ of the Lie group $SO(3)$ are characterized by an integer spin $\ell\in\mathbb{N}_0$, which are related to the $2^{\ell}$-pole term. Here the Casimir $L^2$ has eigenvalue $\ell(\ell+1)$. The irrep $V_{\ell}\equiv\underline{\bf 2\ell\!+\!1}$ has dimension $2\ell\!+\!1$. E.g. $\ell=0$ is a monopole, $\ell=1$ is a dipole, $\ell=2$ is a quadrupole, and so forth. 

*The main point is that potentials for any number of charges can be classified according to above scheme. The corresponding radial $2^{\ell}$-pole solution falls off as $ r^{-(\ell+1)}$. See also related Phys.SE posts here and here.  
A: Think about what it takes to construct each multipole with point charges that all have the same magnitude. 


*

*A monopole takes one point charge.

*The three basic dipoles take two - one positive, one negative, oriented in a combination of the three orthogonal directions.

*The five basic quadrupoles are all constructed from various combinations of two dipoles (four monopoles).

*The seven basic octupoles are all constructed from two quadruoples ($8$ monopoles). 

*etc.


In other words, in order to construct a multipole of a given order you need to arrange the charges in such a way as to cancel all of the lower order multipoles. Thus the easiest way to construct a multipole of order $n$ is to take a multipole of order $n-1$, copy it, offset the copy along an axis of (a)symmetry, then flip all of the charges of the copy. That is, put a $2^{n-1}$-pole next to a $2^{n-1}$-pole in such a way that the $2^{n-1}$-pole of the combination is zero. That requires $2^{n-1} + 2^{n-1}=2^n$ monopoles of equal charge magnitude.
For an explanation of the number of basic types of each multipole, see the second item in @Qmechanic's answer. 
