Solution
= Solution
{c}
The square $Q_2$ and the drawn cube $Q_3$ are planar. For $k\geq4$, the subdivision in part (v) prevents planarity by <Kuratowski's theorem>. One can also check the edge count directly: a triangle-free simple <planar graph> with $n\geq3$ vertices has at most $2n-4$ edges, whereas $Q_k$ has $n=2^k$ and $m=k2^{k-1}\geq2n$. Thus
$$
\boxed{Q_k\text{ is planar exactly for }k=2,3\text{ in the requested range }k\geq2.}
$$