Given that by definition the i-th B-spline of order k is: $$B_{i,k}=w_{i,k}B_{i,k-1}+(1-w_{i+1,k})B_{i+1,k-1}$$
where $w_{j,k}=\frac{x-t_j}{t_{j+k-1}-t_j}$
We can define the spline space as $$S_{k,t}:=\{\sum_i\alpha_iB_{i,k}:\alpha_i \in \mathbb{R}\} $$
The author pointed out that $$\sum_i\alpha_i B_{i,k}=\sum_i(\alpha_i w_{i,k}+\alpha_{i-1}(1-w_{i,k}))B_{i,k-1}$$ but honestly I don't really get how to arrive to this expression, using the recursion I am able to obtain: $$\sum_i\alpha_i B_{i,k}=\sum_i(\alpha_i w_{i,k}B_{i,k-1}+\alpha_{i}(1-w_{i+1,k})B_{i+1,k-1})$$ which is nothing like the desired result.
Any hint will be greatly appreciated.
The two expressions coincide: $$ \begin{split} &\phantom{=}\ \sum_i(\alpha_i w_{i,k}B_{i,k-1}+\alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}) \\ &= \sum_i(\alpha_i w_{i,k}B_{i,k-1}+\alpha_{i-1}(1-w_{i,k})B_{i,k-1}) \\ &= \sum_i(\alpha_i w_{i,k}+\alpha_{i-1}(1-w_{i,k}))B_{i,k-1} . \end{split} $$