I am trying to solve this Martingale exercise which states that : Consider the simple random walk in $\mathbb{Z}$ and the function $f:\mathbb{Z} \to \mathbb{Z},f(k)=k^4$.
Show that : $Lf(k)=6k^2+1 $ and prove that the process $\{M_{n}\}_{n \in N_{0}}$ with $M_{0}=0$ and $$M_{n}=X_{n}^4-X_{n}^2-6\sum_{k=0}^{n-1} X_{k}^2,\quad \forall n \in N$$ is a martingale.
My effort:
For $f(k) =k^4$ , the generator of the chain: $$ Lf(x)=0 \Rightarrow \frac{f(x+1)+f(x-1)}{2} -f(x)=0 $$
becomes:
\begin{align*} Lf(k)& = \frac{f(k+1)+f(k-1)}{2} -f(k)\\ & = \frac{f(k+1)^4+f(k-1)^4}{2} -f(k)^4\\ &= \frac{ (k^4 + 4k^3+6k^2 +4k +1) + (k^4 - 4k^3+6k^2 -4k +1) }{ 2 } -k^4 \\ & = \frac{2k^4 +12k^2+2}{2} -k^4\\ & = k^4 +6k^2 +1 -k^4\\ & = 6k^2 +1 \end{align*}
For the stochastic process $\{M_{n }\}_{n \in \mathbb{N}_{0}}$ with $M_{0}=0$ and $$M_{n}=X_{n}^4-X_{n}^2-6\sum_{k=0}^{n-1} X_{k}^2$$ we have :
\begin{align*} \mathbb{E} \left[ M_{n+1} \mid \mathbb{F}_{n} \right] &= \mathbb{E} \left[ X_{n+1}^4 - X_{n+1}^2 -6 \sum_{k=0}^{n} X_{k}^2\mid \mathbb{F}_{n} \right] \\ &= \mathbb{E} \left[ X_{n+1}^4 \mid \mathbb{F}_{n} \right] - \mathbb{E} \left[ X_{n+1}^2 \mid \mathbb{F}_{n} \right] - 6\sum_{k=0}^{n} \left[ \mathbb{E} \left[ (X_{n} -X_{n-1})^2 \mid \mathbb{F}_{n} \right] \right] \\ & = \mathbb{E} \left[ X_{n}^4 + 4X_{n}^3 X_{n+1} +6X_{n}^2 X_{n+1} + 4X_{n}X_{n+1}^3 +X_{n+1}^4 \mid \mathbb{F}_{n}\right]\\ & - \mathbb{E}\left[ X_{n+1}^{2}\mid \mathbb{F}_{n} \right] - 6\left[ \mathbb{E}[X_{0}^2] + \mathbb{E}\left[X_{n}^{2} - X_{n-1}^{2} + 2 X_{n}^2X_{n-1} \mid \mathbb{F}_{n} \right] \right] \\ &=X_{n}^4 +4X_{n}^3 \underbrace{ \mathbb{E} \left[ X_{n+1} \mid \mathbb{F}_{n} \right]}_{0} +6x_{n}^2 \underbrace{\mathbb{E} \left[ X_{n+1}^2 \mid \mathbb{F}_{n} \right]}_{1} + 4X_{n} \underbrace{\mathbb{E} \left[ X_{n+1}^3 \mid \mathbb{F}_{n} \right]}_{0} + \underbrace{\mathbb{E} \left[ X_{n+1}^4 \mid \mathbb{F}_{n} \right]}_{1} \\ &+\mathbb{E}\left[X_{n}^2 \mid \mathbb{F}_{n} \right] + \underbrace{ \mathbb{E}\left[ X_{n+1}^{2}\mid \mathbb{F}_{n} \right]}_{1}+ \mathbb{E}\left[2X_{n}X_{n+1} \mid \mathbb{F}_{n} \right] -6 \underbrace{\mathbb{E}\left[X_{0}^2 \right]}_{0} + 6 \mathbb{E} \left[X_{n}^2 \mid \mathbb{F}_{n}\right] - \underbrace{6 \mathbb{E}\left[X_{n-1}^2\mid \mathbb{F}_{n} \right]}_{1} \\ &= X_{n}^4 +0+ 6X_{n}^{2} +0+1 + X_{n}^2+1 +2X_{n}\underbrace{\mathbb{E}\left[X_{n+1} \mid \mathbb{F}_{n}\right]}_{0} -0 + 6X_{n}^2 -6 \\ & =X_{n}^4 +6X_{n}^2 +1 + X_{n}^{2} +1 + 6X_{n}^{2} -6 \\ &= X_{n}^4 +X_{n}^2- 4 \end{align*}
I have made something very wrong here and i don't know what.Any help ?