}{ (1-q)^n (1-q^{-1}) }   \bigg)$$

$$= {\rm CT}_q \bigg(  \sum_{j=0}^n \binom nj (-1)^{j+1}  q^{rtj+j}  \frac {  ( q^{-kt+t} - q^{-kt } )   }{ (1-q)^{n+1} }   \bigg)$$

$$= {\rm CT}_q \bigg(  \sum_{j=0}^n \sum_{i\geq0}  \binom nj \binom{n+i}{n}  (-1)^{j+1}  q^{rtj+j+i}   ( q^{-kt+t} - q^{-kt } )  \bigg)$$

$$= \sum_{j=0}^n  \binom nj   (-1)^{j+1}  \left( \binom{n+kt+t-(rt+1)j}{n} - \binom{n+kt-(rt+1)j}{n}  \right)$$ with the (unusual) convention that $\binom nk=0$ when $n<0$. With this convention, it is not clear that we have a polynomial in $t$. But we can improve the formula by keeping only some of the indices $j$, those appearing in the announced formula. Indeed both formulas are equal for large $t$, hence for every $t$ since these are polynomials.

The Ehrhart polynomial of $\mathcal{B}^{(r)}_{n,k}$ is: $$\begin{aligned}
      \sum_{j=0}^{\lfloor (k-1)/r \rfloor} (-1)^{j+1}  \tbinom nj \tbinom{n-rtj-j+kt-t-1}{n} - 
                      \sum_{j=0}^{\lfloor k/r \rfloor  } (-1)^{j+1}  \tbinom nj \tbinom{n-rtj-j+kt-1}{n}.\end{aligned}$$

This is similar to the previous proposition: $$\# \big( \mathbb{Z}^n \cap t\mathcal{B}^{(r)}_{n,k} \big)  
                   = \# \big\{ v \in \{0,1,\dots,rt\}^n  \, : \,  kt-t \leq \sum v_i < kt \big\}$$

$$=  {\rm CT}_q \bigg(  [rt+1]_q^n  ( [kt]_{q^{-1}} - [kt-t]_{q^{-1}} )  \bigg)$$

$$= {\rm CT}_q \bigg(  \frac{ (1-q^{rt+1})^n  ( q^{-kt+t} - q^{-kt } )   }{ (1-q)^n (1-q^{-1}) }   \bigg)$$

$$= {\rm CT}_q \bigg(  \sum_{j=0}^n \binom nj 