Solution
= Solution
Expansion of $\det(xI-A_n)$ along the last row gives
$$
D_n=(x-\alpha_n)D_{n-1}-\beta_nD_{n-2},
$$
with $D_0=1$, $D_1=x-\alpha_1$. This is exactly the recurrence defining $P_n$, so induction gives $P_n=\det(xI-A_n)$.
= Solution
Expansion of $\det(xI-A_n)$ along the last row gives
$$
D_n=(x-\alpha_n)D_{n-1}-\beta_nD_{n-2},
$$
with $D_0=1$, $D_1=x-\alpha_1$. This is exactly the recurrence defining $P_n$, so induction gives $P_n=\det(xI-A_n)$.