The optimal third-order method is $$\begin{align*} \alpha_1 = & \frac{1}{2k+1)(k-1)^2} & \alpha_{k} = & \frac{k^2(2k-3)}{(k-1)^2(2k+1)} \\ \beta_k = & \frac{k^2}{(k-1)(2k+1)} & \beta_{k+1} = & \frac{k}{2k+1)} \\ \end{align*} $$ and has time step coefficient $\frac{2k-3}{k-1}$.