jmc

We have $n=50a-1$ for some integer $a$, so $n\equiv -1(\mathrm{mod}50)$. Therefore, $$\begin{array}{rl}{n}^2+2n+3& \equiv (-1{)}^2+2(-1)+3\\ & \equiv 1-2+3\\ & \equiv 2(\mathrm{mod}50).\end{array}$$The remainder when $n^2+2n+3$ is divided by $50$ is $\boxed{2}$.