It is a family of numerical methods to solve ODEs. The more famous is the RK4 (4th order).
$$ \begin{array}{l} y_{n+1}=y_n+h \frac{1}{6}\left(k_1+2 k_2+2 k_3+k_4\right) \\ t_{n+1}=t_n+h \end{array} $$where
$$ \begin{array}{l} k_1=f\left(t_n, y_n\right) \\ k_2=f\left(t_n+\frac{h}{2}, y_n+h \frac{k_1}{2}\right), \\ k_3=f\left(t_n+\frac{h}{2}, y_n+h \frac{k_2}{2}\right), \\ k_4=f\left(t_n+h, y_n+h k_3\right) . \end{array} $$Here $y_{n+1}$ is the RK4 approximation of $y(t_{n+1})$, and the next value ($y_{n+1}$) is determined by the present value ($y_{n}$) plus the weighted average of four increments, where each increment is the product of the size of the interval, $h$, and an estimated slope specified by function f on the right-hand side of the differential equation.
Short answer: you can think of $v:=\frac{1}{6}\left(k_1+2 k_2+2 k_3+k_4\right)$ as a weighted average slope, so
$$ y_{n+1}=y_n+h v_n $$is nothing but a kind of Euler method but with an improved slope. Instead of taking the slope at $(x_n,y_n)$ you take several measures ($k_i$) in the interval $(x_n,x_{n+1})$. Why these weights exactly? This requires a long answer.
Long answer: see this great answer. To summarize, the exact value of $y_{n+1}$ is $y_n+\int_{x_n}^{x_{n+1}}f(x,y)dx$ so you should approximate this integral numerically (because you don't know $y(x)$). When you use Simpson's rule you obtain the formula you have presented.
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Author of the notes: Antonio J. Pan-Collantes
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