Backward Differential Equation with binomial tree I'm trying to understand/solve the following question but I honestly don't know what it's even asking about.  I've included my attempt following the picture of the question.

I would approximate the first derivative of $F$ at $(x,t)$ by $$F_x(x,t) \approx \frac{F(x+dx, t+2dt) - F(x, t+2dt)}{dx} \equiv \Delta F(x, t)$$  giving rise to $$\frac{d^2}{dx^2}F(x,t) \approx \frac{\Delta F(x + dx, t) - \Delta F(x,t)}{dx} = \frac{F(x+2dx, t+2dt) - 2F(x+dx, t+2dt) + F(x,t+2dt)}{(dx)^2} = \frac{\Delta F(x + dx, t) - \Delta F(x,t)}{dx} = \frac{F(x+2dx, t+2dt) - 2F(x+dx, t+2dt) + F(x,t+2dt)}{\sigma^2 dt}$$
This is obviously different from what's given in the question statement, as the terms and the factor of 4 is missing.  I'm completely lost.  Any help would be massively appreciated!

A classical outset for the presented finite difference to approximate the second order derivative is the following reasoning:
"Decompose" $\partial_{xx} F$ into $\partial_x \underbrace{\partial_x F}_{=:u}$ and use the central difference to approximate the "outer" derivative:
$$\partial_x u \approx \frac{u(x + d x ) - u(x - dx)}{2dx}$$
I believe this is what the first arrows in the image could describe.
Then, you use again central differences to be able to express $u(x \pm dx)$ in terms of $F$:
$$ u(x \pm dx) = \partial_x F \Big \vert_{x \pm dx} \approx \frac{F(x \pm d x + dx ) - F(x \pm dx - dx)}{2dx} $$
Thus,
\begin{align}
\partial_{xx} F \approx & \frac{\frac{F(x + 2 dx ) - F(x + dx - dx)}{2 dx} - \frac{F(x - dx + dx ) - F(x -dx -dx)}{2dx}}{2dx} \\
=& \frac{F(x + 2dx) - 2 F(x) - 2 F(x - dx)}{4 dx^2} \\
\overset{dx = \sigma \sqrt{dt}}{=} &\frac{F(x + 2dx) - 2 F(x) - 2 F(x - dx)}{4 \sigma^2 dt} \\
\end{align}
Note that these steps hold for any time $t$, so in particular also for $t + 2 dt$.
