Let $N_j$, $j=-k,\ldots , m-1$ the normalized B-splines of the set of nodes $x_0, \ldots , x_m$ of degree $k$.
Show that $$\sum_{j=-k}^{m-1}N_j(x)=1 \ \text{ for all } x\in [x_0, x_m]$$
Do we use for that the below formula? $$ N_{i,0,\tau }(u)=\begin{cases}1,&u\in \left[\tau _{i},\tau _{i+1}\right)\\0,&{\text{else}}\end{cases} \\ N_{i,p,\tau }(u)={\frac {u-\tau _{i}}{\tau _{i+p}-\tau _{i}}}\,N_{i,p-1,\tau }(u)\;+\;{\frac {\tau _{i+p+1}-u}{\tau _{i+p+1}-\tau _{i+1}}}\,N_{i+1,p-1,\tau }(u) \text{ for } p>0$$
Or do we use an other formula to calculate the sum?
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EDIT:
In the notes I found also a formula with divided differences, which is \begin{align*}&N_j(x)=(x_{j+k+1}-x_j)B_j(x) \\ \text{ with } \ B_j&=(\cdot -x)_+^k[x_jx_{j+1}x_{j+2}\ldots x_{j+k+1}]\\ & =\frac{(\cdot -x)_+^k[x_{j+1}x_{j+2}\ldots x_{j+k+1}]-(\cdot -x)_+^k[x_jx_{j+1}x_{j+2}\ldots x_{j+k}]}{x_{j+k+1}-x_j}\end{align*} where $(x)_+^k=\begin{cases}x^k & \text{ if } x\geq 0 \\ 0 & \text{ if } x<0\end{cases}$.
Then we have \begin{align*}&\sum_{j=-k}^{m-1}N_j(x)=\sum_{j=-k}^{m-1}(x_{j+k+1}-x_j)B_j(x)\\ & =\sum_{j=-k}^{m-1}(x_{j+k+1}-x_j)\frac{(\cdot -x)_+^k[x_{j+1}x_{j+2}\ldots x_{j+k+1}]-(\cdot -x)_+^k[x_jx_{j+1}x_{j+2}\ldots x_{j+k}]}{x_{j+k+1}-x_j}\\ & =\sum_{j=-k}^{m-1}\left ((\cdot -x)_+^k[x_{j+1}x_{j+2}\ldots x_{j+k+1}]-(\cdot -x)_+^k[x_jx_{j+1}x_{j+2}\ldots x_{j+k}]\right )\end{align*} How can we continue?
Yes. you use those formulas. In fact, your course probably gave those formulas as the definitions of b-spline basis functions, so, if you don't know any other facts about them, there's no other place to start.
In case it's not obvious, you use induction on the degree, $p$.