Let $i : \mathbf P^1 \to \mathbf P^2$ be the second Veronese embedding. Clearly, $i\_\star \mathcal O\_{\mathbf P^1}$ has a locally free resolution of the form


$$
0 \to \mathcal O\_{\mathbf P^2} (-2) \to \mathcal O\_{\mathbf P^2} \to i\_\star \mathcal O\_{\mathbf P^1} \to 0
$$


More generally, for any integer $n$, we have $\mathcal O\_{\mathbf P^1} (2n) \cong \mathcal O\_{ \mathbf P^1} \otimes i^\star \mathcal O\_{\mathbf P^2} (n)$, so $i\_\star \mathcal O\_{\mathbf P^1} (2n) $ has a resolution


$$
0 \to \mathcal O\_{\mathbf P^2} (n-2) \to \mathcal O\_{\mathbf P^2} (n) \to i\_\star \mathcal O\_{\mathbf P^1}(2n) \to 0
$$


My question is: how can I construct a locally free resolution for $i\_\star O\_{\mathbf P^1} (2n+1)$?


(I should add that it would make my life easier if the sheaves in this resolution are direct sums of invertible sheaves on $\mathbf P^2$.)



Veronese embeddings and locally free resolutions
The resolution looks like,
$$0\to\mathcal{O}\_{\mathbb{P}^2}(n-1)^{\oplus 2}\to \mathcal{O}\_{\mathbb{P}^2}(n)^{\oplus 2}\to i\_\*\mathcal{O}\_{\mathbb{P}^1}(2n+1)\to 0,$$ with the determinant of the $2\times 2$ matrix appearing in the left most map is just the equation of the quadric defining $i(\mathbb{P}^1)$.


