Let $S_n$ be defined as $\frac{1}{n}\sum_{t=1}^{t=n} [px_t^2 - (p+q)x_t]$ where $x_t = 1-(1-p-q)^t$. We want to find the conditions on $p$ and $q$ such that $S_n$ is monotonically decreasing for all $n$. $0 < p,q < 1$ and $-1 < 1-p-q < 1$.
Note: I tried to prove this by taking the difference between consecutive terms of the summand i.e. $[px_t^2 - (p+q)x_t] - [px_{t+1}^2 - (p+q)x_{t+1}]]$ and trying to show that it is greater than 0 but I am not able to make progress on this.
Since $x_0=0$, it will be convenient to do summation starting from $t=0$.
Denoting $r:=1-p-q$, we have \begin{split} S_n &= \frac1n\sum_{t=0}^n \left(pr^{2t} + (q-p)r^{t} - q\right) \\ &=\frac{p}{1-r^2} \frac{1-r^{2(n+1)}}{n} + \frac{q-p}{1-r} \frac{1-r^{n+1}}{n} - \frac{n+1}{n}q. \end{split}
Given that $|r|<1$, it can be easily verified that $$\frac{\partial S_n}{\partial n} = \frac{(p - (r + 1) q)r}{n^2(1-r^2)} + O(r^n)$$ as $n$ grows.
Hence, $S_n$ decreases for large enough $n$ iff $p-(r+1)q\lessgtr 0$ or $$p\lessgtr\frac{(2-q)q}{1+q},$$ where the inequality sign is "$<$" if $r>0$, and "$>$" if $r<0$. In other words, either of the following two conditions works: $$p < \min\left\{\frac{(2-q)q}{1+q},1-q\right\}$$ or $$p > \max\left\{\frac{(2-q)q}{1+q},1-q\right\}.$$
