I am trying to show that if we let $\{Z_{n},n\geq 1\}$ independent and identically distributed random variables with $P\left[Z_{1}=1 \right]=p=1-P\left[Z_{1}=0\right]$. Let $T_{n}=\sum_{j=1}^{n}Z_{j}$ be a random walk on $\{0,1,2,..\}$ that can either make one step to the right or stay in its current state and that it starts at state 0. Then, I am trying to show that for $n\geq 0$ and $1\geq j \geq n+1$, $P[T_{n+1}=j]=pP[T_{n}=j-1]+(1-p)P[T_{n}=j]$.
I already show this using Mathematical Induction but I want to do this using the law of total probability. Can someone help with this?
$$P(T_{n+1}=j)= P(T_{n+1}=j|Z_1=1)P(Z_1=1)+P(T_{n+1}=j|Z_1=0)P(Z_1=0)$$ $$=pP(T_n=j-1)+(1-p)P(T_n=j)$$ where the last step follows from the fact that the sequences $( \sum\limits_{k=1}^{n} Z_k)$ and $( \sum\limits_{k=1}^{n} Z_{k+1})$ have the same distribution (because $(Z_j)$ is i.i.d.).