So I have been reading about the Irwin-Hall distribution online, it is a sum of uniform distributions on $[0,1]$, and it seems very interesting:
On the Wikipedia article above they derive pdf for the special cases n = 1,2,3,4, and 5. n = 1 is trivial, for n = 2 we are drawing points from a square $U_1 \times U_2$. Then we compute the probability of picking points under a line in that square, take its derivative and then derive the triangular distribution. For n = 3, we are drawing points from the cube $U_1 \times U_2 \times U_3$ and compute the probability of picking points under a plane and can intuitively see the parabolic distribution (since the volume under the plane on the cube will be of third degree, then we takes its derivative to get pdf).
In general we see the pdf will have n pieces with each piece of degree k-1. My question here is deriving the formulas for the n = 3,4,5 case don't seem easy as they are shown in the Wikipedia article, what would be the approach to get the equations? Secondly, intuitively we would think due to the central limit theorem that this distribution approaches the normal distribution, but they give a general version of the pdf in the Wikipedia article, and I don't see how that is going to converge to the normal distribution.
For n = 3, I know how to get pdf when $x \in [0,1]$ and $x \in [2,3]$, but not sure about when $x \in [1,2]$.