= Solution
We first derive the recurrence needed for the <partition of unity>. Denote an order-$r$ partition-normalized <spline> by $N_{i,r}$. For $r\geq2$, set $h_t(u)=(u-t)_+^{r-2}$, so $(u-t)h_t(u)=(u-t)_+^{r-1}$. The elementary product identities for <divided differences> are
$$
\begin{aligned}
[u_0,\ldots,u_s](u h(u))&=u_0[u_0,\ldots,u_s]h+[u_1,\ldots,u_s]h\\
&=u_s[u_0,\ldots,u_s]h+[u_0,\ldots,u_{s-1}]h.
\end{aligned}
$$
They follow from the <Leibniz rule for divided differences> because the divided differences of the linear factor vanish above first order. Use the first identity on $t_i,\ldots,t_{i+r-1}$ and the second on $t_{i+1},\ldots,t_{i+r}$. Subtracting cancels the common middle-block term. The defining divided-difference recurrence for $N_{i,r}$ then gives
$$
\boxed{N_{i,r}(t)=\frac{t-t_i}{t_{i+r-1}-t_i}N_{i,r-1}(t)
+\frac{t_{i+r}-t}{t_{i+r}-t_{i+1}}N_{i+1,r-1}(t).}
$$
This is the <Cox-de Boor recurrence>, derived here from the specified normalization rather than assumed as a partition theorem.
Extend the distinct knot sequence in both directions to a strictly increasing sequence $(t_i)_{i\in\mathbb Z}$ tending to $\pm\infty$. For order one, $N_{i,1}=\mathbf1_{[t_i,t_{i+1})}$, so the full sum equals one. All higher-order <splines> are supported on $[t_i,t_{i+r}]$, making the following sums locally finite. Reindexing the second term of the recurrence yields
$$
\begin{aligned}
\sum_{i\in\mathbb Z}N_{i,r}(t)
&=\sum_{j\in\mathbb Z}\left[\frac{t-t_j}{t_{j+r-1}-t_j}
+\frac{t_{j+r-1}-t}{t_{j+r-1}-t_j}\right]N_{j,r-1}(t)\\
&=\sum_{j\in\mathbb Z}N_{j,r-1}(t).
\end{aligned}
$$
Induction proves that the full sum is one for every order. The same recurrence gives nonnegativity: each <coefficient> is nonnegative wherever its associated lower-order <spline> is nonzero. Removing <splines> therefore gives the <subpartition of unity for B-splines> on the whole line.
For $t_k<t<t_{n+1}$, a <spline> with index $i\leq0$ ends no later than $t_k$, and one with $i\geq n+1$ starts no earlier than $t_{n+1}$. They all vanish, leaving exactly the indices $1,\ldots,n$. For $k\geq2$ the <splines> are continuous and vanish at their <support> endpoints, so the equality extends to the closed basic knot interval:
$$
\boxed{\sum_{i=1}^nN_i(t)=1,\qquad t_k\leq t\leq t_{n+1}.}
$$
If order one is allowed, the standard half-open indicator convention gives equality on $[t_1,t_{n+1})$; to include its final endpoint literally, assign the left-limit value there. This harmless endpoint convention does not change integrals or $L^\infty$ <norms>. For the continuous <splines> of order at least two no such convention is necessary.
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