We define the boolean lattice graph $BL_n$ with $n\ge 1$ as the graph with vertex set the power set of $\{1,\dots,n\}$ and two vertices are adjacent iff their symmetric difference is exactly one.
I can show that the amount of edges of $BL_n=n2^{n-1}$, by altering each set by one and diving by two to account for the double counts. However, I would like to prove it via induction and struggle to see the induction step.
For $n=1$ the statement is clear. Assuming it is true for $BL_{n-1}$ we would like to prove it for $BL_n$.
I've tried to construct $BL_n$ from $BL_{n-1}$, where we add $2^{n-1}$ vertices and now need to find the amount of edges we add. Clearly, I can take each vertex from $BL_{n-1}$ and add the new element $n$. By this I've created all the connection between $BL_{n-1}$ and $BL_n$. However, there are missing edges, with vertices both only in $BL_{n}$. Formally, I would like to show
$$ n2^{n-1} = (n-1)2^{n-2}+x$$
where $x$ are the added edges. Solving for $x$, leads to $$x=2^{n-2}(n+1)$$
I can't see how this is derived geometrically. Any hint would be much appreciated.
$BL_{n+1}$ will consist of two copies of $BL_n$, one whose sets do not contain $n+1$ and one whose sets do contain $n+1$. There are $ 2^{n}$ edges between these two copies. So there are \begin{eqnarray*} 2 \times n 2^{n-1} + 2^n = (n+1) 2^n \end{eqnarray*} edges in $BL_{n+1}$.