Show $E[Y_{n+1}|F_n]=Y_n+\frac{1}{2^{n+1}}E[X_0]$ and show $Z_n$ is a martingale

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Problem:

Let $(X_n)_{n\geq 0}$ be a sequence of i.i.d. distributed random variables on $\{0,1,...,k\}, k \geq 2.$ Let $F_n=\sigma(X_1,...,X_n)$ (that is $F_n$ is a filtration) and define:

$$Y_n=\displaystyle\sum_{i=0}^n \frac{1}{k^i} X_i, n\geq 1$$

$1.$ Show that $E[Y_{n+1}|F_n]=Y_n+\frac{1}{2^{n+1}} E[X_0]$

Define $$Z_n=Y_n-A_n, \: \: A_n=\displaystyle\sum_{i=0}^n \frac{1}{k^i}E[X_0], \: n\geq 0$$

$2.$ Show that $Z_n$ is a martingale wrt $(F_n)_{n\geq 0}$

Attempt:

$1.$

$$E[Y_{n+1}|F_n]=E[\displaystyle\sum_{i=0}^n \frac{1}{k^i} X_i+\frac{1}{k^{n+1}}X_{n+1}|F_n]=$$ $$=E[\displaystyle\sum_{i=0}^n \frac{1}{k^i} X_i|F_n]+E[\frac{1}{k^{n+1}}X_{n+1}|F_n]=$$ $$=\displaystyle\sum_{i=0}^n \frac{1}{k^i} X_i+\frac{1}{k^{n+1}}E[X_{n+1}]=Y_n+\frac{1}{k^{n+1}}E[X_{0}]$$

My argument from here is just that i can pick $k=2$ and that's the end of it (?)

$$=Y_n+\frac{1}{2^{n+1}}E[X_{0}]$$

(Throughout the calculation i've used that $X_n$ is $F_n$-measurable and $X_{n+1}$ and $F_n$ is independent.)

Is my argumentation wrong?

$2.$

It's only the first of the three conditions i need help with:

I want to show that $E[|Z_n|]<\infty$

Well, my attempt is using proof by induction, how does that sound?

The only problem is just that how can i be sure that $E[X_0]$ is well-defined?