= Solution
Write $N_{i,k}$ for the order-$k$ <B-spline>. At order one, $N_{i,1}=\mathbf1_{[t_i,t_{i+1})}$, and the finite sum is one on its basic knot interval. For larger order, the <Cox-de Boor recurrence> is
$$
N_{i,k}(t)=\frac{t-t_i}{t_{i+k-1}-t_i}N_{i,k-1}(t)+\frac{t_{i+k}-t}{t_{i+k}-t_{i+1}}N_{i+1,k-1}(t),
$$
with a zero-denominator term defined as zero. Sum over $i=1,\ldots,n$. For each interior lower-order index $j=2,\ldots,n$, the combined coefficient is
$$
\frac{t-t_j}{t_{j+k-1}-t_j}+\frac{t_{j+k-1}-t}{t_{j+k-1}-t_j}=1.
$$
The two unmatched lower-order terms have supports $[t_1,t_k]$ and $[t_{n+1},t_{n+k}]$, respectively, and vanish on the open basic interval $(t_k,t_{n+1})$. Induction therefore gives
$$
\boxed{\sum_{i=1}^nN_{i,k}(t)=1,\qquad t_k\le t\le t_{n+1}}.
$$
For simple knots and $k\ge2$, <continuity> extends the identity to both endpoints. In the order-one case and at admissible repeated endpoint knots, use the one-sided value from the basic interval. Extending the knot sequence in both directions also proves the <subpartition of unity for B-splines>, $0\le\sum_iN_i\le1$ everywhere: the finite sum is a subset of a nonnegative full partition. That weaker global inequality will be needed in Question 7(a).
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