$\mathbf{Premise:}$ We wish to find the equivalent resistance between $(0,0,0,...,0)$ and $(1,1,1,...,1)$, where resistors of resistance $R$ are placed on the edges of the cube
$\mathbf{Definitions:}$
The set of all vertices of the unit n-cube is $V=\{(0,0,0,...,0),(0,0,0,...,1),...,(1,1,1,...,1)\}$
The "order" of a vertex be the shortest distance to the origin along the edges of the cube.
The set of edges connecting vertices of order $k-1$ and $k$ is $E_{k}$
$\mathbf{Attempt:}$
Noticing that adjacent vertices differ by only one "bit" eg:$(0,0,0,...,0)$ and $(0,0,0,...,1)$ are connected by an edge, it follows that the number of edges (resistors) connecting vertices of order $k-1$ and $k$ is,$$e_{k}=\binom{n}{k}$$
Therefore, the equivalent resistance for $E_k$ is: $$\frac{1}{R_{k}}=\frac{1}{R}\binom{n}{k}$$
The equivalent resistance of the n-cube is : $$R_{eq}=R\sum_{k=1}^{n-1}\frac{1}{\binom{n}{k}}$$ I have no clue as to how to compute the sum, if a closed form exists, and Wolfram is showing something weird.
