= Solution
Put $n=T-t$ and define the finite <geometric series>
$$
A_0=0,\qquad A_j=\sum_{k=0}^{j-1}\beta^k
=\begin{cases}(1-\beta^j)/(1-\beta),&\beta\ne1,\\j,&\beta=1.\end{cases}
$$
Starting the <autoregressive process of order one> at time $t$ gives $r_{t+k}=\beta^kr_t+\sum_{j=1}^k\beta^{k-j}\xi_{t+j}$. Consequently
$$
\sum_{k=1}^n r_{t+k}=\beta A_nr_t+\sum_{j=1}^nA_{n-j+1}\xi_{t+j},\qquad
\frac{B_t}{B_T}=\exp\left(-\sum_{k=1}^nr_{t+k}\right).
$$
The future innovations are <independent and identically distributed random variables>, independent of $\mathcal F_t$. Factorizing their exponential <conditional expectation> and using the <cumulant-generating function> $K$ yields
$$
P(t,T)=e^{-\beta A_nr_t}\prod_{j=1}^n\mathbb E[e^{-A_{n-j+1}\xi_1}]
=\exp\left(-\beta A_nr_t+\sum_{j=1}^nK(-A_j)\right).
$$
Thus the <exponential-affine bond pricing> coefficients are
$$
\boxed{Q(t,T)=-\beta A_{T-t},\qquad R(t,T)=\sum_{j=1}^{T-t}K(-A_j).}
$$
They vanish at maturity, giving $P(T,T)=1$. Everywhere finiteness of $K$ ensures that each <zero-coupon bond> price is positive and finite, including when $\beta$ is zero, negative, or one.
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