= Solution
{c}
Write $v=\sum_{i=1}^{k+1}\alpha_i u_i$. Since $v\notin\langle u_1,\ldots,u_k\rangle$, necessarily $\alpha_{k+1}\ne0$. Rearranging gives
$$
u_{k+1}=\frac1{\alpha_{k+1}}v-\sum_{i=1}^k\frac{\alpha_i}{\alpha_{k+1}}u_i
\in\langle u_1,\ldots,u_k,v\rangle.
$$
All generators of $\langle u_1,\ldots,u_{k+1}\rangle$ therefore belong to $\langle u_1,\ldots,u_k,v\rangle$, proving one <set inclusion>. Conversely, $u_1,\ldots,u_k,v$ all belong to the <linear subspace> $\langle u_1,\ldots,u_{k+1}\rangle$, proving the other <set inclusion>. Thus
$$
\boxed{\langle u_1,\ldots,u_k,v\rangle=\langle u_1,\ldots,u_{k+1}\rangle.}
$$
This <linear span> exchange does not require <linear independence> of the original <vectors>.
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