How many node does the final B-tree have?
Every node contains between $\lceil(m/2)\rceil-1$ and $m-1$ keys (where m is the degree), so we can say that every node has between $\lceil(m/2)\rceil$ and $m$ children.
If we imagine to construct a minimun nodes b-tree, we'll have:
$n = 1 + 2  + 2\lceil m/2\rceil + 2\lceil m/2\rceil^2 + ... + 2\lceil m/2\rceil^{h-2}$where every addend is the number of nodes for every level, from the root (that can contains even one key), to the leaves level, and 
where h is the height of the b-tree; 
If you consider the series, you have $n=1+2 \frac{ \lceil m/2\rceil^{h-1}-1}{\lceil m/2\rceil-1}$.
Instead, if you consider the maximum number of nodes, you will have:
$n= 1+m+m^2+...+m^{h-1} =  \frac{m^h-1}{m-1}$.
where at the same way, every addend is the number of nodes for every level, and 
where h is the height of the b-tree; 
You can easily find the result if you substitute 2 instead of m in the formula.