Proving a given scheme is consistent with PDE 
Exercise $1$:
We consider the following scheme
$$U_{j,n+1} = U_{j,n} + r\delta_{x}^{2}U_{j,n} + 2r\delta_{x}^{2}U_{j,n+1},$$ where $r = \frac{\Delta t}{\Delta x^{2}}$.
$1$. Show that this scheme is consistent with the PDE $u_{t} = 3u_{xx}$.

I can't seem to prove this.  I know that $u_{t}$ can be approximated as
$$\frac{F_{t}}{ \Delta t}  =\frac{U_{j}^{n+1} - U _{j}^{n}}{ \Delta}.$$
However, I am not to sure about $u_{xx}$, should I Taylor expand?
 A: Notice that you can rewrite the scheme as $$\frac{U_j^{n + 1} - U_j^n}{\Delta t} = \frac{U_{j + 1}^n - 2U_j^n + U_{j - 1}^n}{\Delta x^2} + 2\frac{U_{j + 1}^{n+1} - 2U_j^{n+1} + U_{j - 1}^{n+1}}{\Delta x^2}.$$
Taylor's expansion on the LHS gives $U_t$ plus terms that go to zero as $\Delta t \to 0$. For the RHS, the analysis of the first term is straightforward: you get $U_{xx}$ plus terms that go to zero as $\Delta x \to 0$. For the second term use Taylor once again: 
$$
\begin{align}
U_{j + 1}^{n +1} = &\ U_j^n +\Delta x \partial_xU_j^n + \Delta t \partial_tU_j^n + \frac{1}{2}(\Delta x^2\partial_{xx}U_j^n + 2\Delta t \Delta x\partial_x\partial_tU_j^n + \Delta t^2\partial_{tt}U_j^n) + \mathcal{O}(\Delta x^2 + \Delta t^2) \\
U_{j - 1}^{n +1} = &\ U_j^n -\Delta x \partial_xU_j^n + \Delta t \partial_t U_j^n + \frac{1}{2}(\Delta x^2\partial_{xx}U_j^n - 2\Delta t \Delta x\partial_x \partial_tU_j^n + \Delta t^2\partial_{tt}U_j^n) + \mathcal{O}(\Delta x^2 + \Delta t^2) \\
U_j^{n + 1} = &\ U_j^n + \Delta t\partial_tU_j^n + \frac{1}{2}\Delta t^2\partial_{tt}U_j^n + \mathcal{O}(\Delta t^2).
\end{align}
$$
Then the terms $U_j^n,\partial_xU_j^n,\partial_tU_j^n,\partial_x\partial_tU_j^n, \partial_{tt}U_j^n$ cancel and you are left with two copies of $U_{xx}$.
