\begin{align*} \begin{array}{c|cc} 0 &0 &0 \\ 1 & \frac{1}{2} & \frac{1}{2} \\ \hline & \frac{1}{2} & \frac{1}{2} \end{array} \end{align*}
How is the trapezoidal method derived from the implicit RK tableau above?
For implicit RK methods, I have that:
\begin{align}
k_j &= y_n + h \sum_{i=1}^s a_{ji}f(t_n + c_ih,k_i) & k=j,...,s\\
y_{n+1} &= y_n +h \sum_{j=1}^s b_j f(t_n + c_jh, k_j)\\
\end{align}
So that from the tableau:
\begin{align}
k_1 &= y_n \\
k_2 &= y_n + \frac{h}{2} \left ( f(t_n, k_1) + f(t_n +h, k_2) \right) \\
&= y_n + \frac{h}{2} \left ( f(t_n, y_n) + f(t_n +h, k_2) \right) \\
\end{align}
which gives
\begin{equation}
y_{n+1} = y_n + \frac{h}{2} \left (f(t_n,y_n) + f \left ( t_n + h, y_n + \frac{h}{2} \left ( f(t_n, y_n) + f(t_n +h, k_2 )\right) \right ) \right ) \tag{1} \label{eq:1}
\end{equation}
How do I go from the equation in \eqref{eq:1} to derive the form below?:
\begin{align}
y_{n+1} &= y_n + \frac{h}{2} \left ( f(t_n,y_n) + f(t_{n+1},y_{n+1}) \right )
\end{align}
Specifically, how do I remove the $k_2$ in (\ref{eq:1})?
By visual inspection, I can observe that what $k_2$ brought into the inner $f$ is equivalent to $y_{n+1}$ as formulated in \eqref{eq:1}: \begin{equation} y_{n+1} = y_n + \frac{h}{2} \left (f(t_n,y_n) + f \left ( t_n + h, \underbrace{y_n + \frac{h}{2} \left ( f(t_n, y_n) + f(t_n +h, k_2 )\right)}_{y_{n+1}} \right ) \right ) \tag{2} \label{eq:2} \end{equation}
But how is this obtained through algebraic manipulations?