Help understanding the solution to a problem regarding kinetic energy of a group of point charges The problem provided by my professor goes as follows:
"Now consider a situation in which all charges are equal to q and they simultaneously become "unglued". What speed will each charge have when a hexagonal configuration has doubled in size (each side has a length).
The work done was found to be $$U=\frac{kq^2}a\left(\frac{5}2+\frac{2}{\sqrt2}\right)$$
So far I have done:
$$\Delta K=-\Delta U$$
$$K_f=U_i-U_f$$
$$6\cdot\left(\frac{1}2mv^2\right)= k\frac{q^2}a\left(\frac{5}2+\frac{2}{\sqrt3}\right)-\frac{kq^2}{2a}\left(\frac{5}2+\frac{2}{\sqrt3}\right)$$
However, the solution is:
$$6\cdot\left(\frac12mv^2\right)=k\frac{q^2}a\left(\frac{15}2+\frac{6}{\sqrt2}\right)-\frac{kq^2}{2a}\left(\frac{15}2+\frac{6}{\sqrt3}\right)$$
Can anyone explain why?
 A: Notice that what you need to do to calculate the whole potential energy is to just count the pair of charges and add the potential between them. What I mean by that is the following:
On a hexagon you have 6 vertices with length say $a$, which corresponds to a potential $\frac{kq^2}{a}$ each, 6 diagonals with length $\sqrt 3 a$, which corresponds to a potential $\frac{kq^2}{\sqrt3 a}$ each and finally 3 diagonals with length $2a$, which corresponds to a potential $\frac{kq^2}{2 a}$ each. What you do is to sum it up ie.
$$U=\frac{kq^2}{a}\left(6 + 3 \cdot \frac{1}{2} + 6 \cdot \frac{1}{\sqrt 3}\right)= \frac{kq^2}{a}\left(\frac{15}{2}+\frac{6}{\sqrt 3}\right)$$
You know how to proceed forward from this expression.
By doing this way you avoid making mistakes because of doing too much algebra. Note that this procedure also works for 3D.
A: I think $U$ must be the potential energy to bring the last particle in from infinity. Thus $U/5$ is the average pairwise  potential between this last bead and each of its neighbors. By symmetry though, each bead has this same average pairwise potential with the other beads. The total potential of the configuration is half the sum of the pairwise potential of all neighbors. Since there are six beads and each has 5 neighbors, the initial total potential energy is $\frac{1}{2}6*5*U/5=3U$.
Then the final potential energy after you expand everything by a factor of 2 is $3U/2$, since $U$ scales with distance${}^{-1}$. 
$3U/2$ is exactly what you have in the last line (modulo typos) once you realize $\dfrac{kq^2}{a} - \dfrac{kq^2}{2a} = \dfrac{kq^2}{2a}.$
