The divisor $D_1$ is a fiber of the ruling and has $D_1^2=0$. Hence $(kD_1)^2=0$ for every $k$. An ample divisor on a complete surface has positive self-intersection, so no $\mathcal O(kD_1)$ with $k\geq0$ is ample. These results are summarized by <multiples of a fiber on the first Hirzebruch surface>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 152 / 3 / c / Solution

For a toric fiber divisor $F$ on $\mathbb F_1$,
$$
h^0(\mathbb F_1,\mathcal O(kF))=
\begin{cases}k+1,&k\geq0,\\0,&k<0.
\end{cases}
$$
The bundles $\mathcal O(kF)$ are basepoint-free for $k\geq0$ but never ample, since $F^2=0$.

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 Incoming links: Multiples of a fiber on the first Hirzebruch surface