= Solution
Let $d_n(x)=\det(xI-A_n)$, and set $d_0=1$, $d_1=x-a_1$. Expansion of the tridiagonal <determinant> along its last row gives
$$
d_n=(x-a_n)d_{n-1}-b_n^2d_{n-2}.
$$
For the second term, the last-row entry is $-b_n$, its cofactor has negative sign, and the remaining minor has last-column entry $-b_n$ multiplying $d_{n-2}$; their product is $-b_n^2d_{n-2}$. Thus $d_n$ and $p_n$ have identical initial values and recurrence, and induction proves
$$
\boxed{p_n(x)=\det(xI-A_n).}
$$
The <eigenvalues> of the <Jacobi matrix> $A_n$ are precisely the roots of this <characteristic polynomial>. Part(a) therefore gives \b[all $n$ eigenvalues simple and lying strictly in $(a,b)$]. This conclusion follows from orthogonality as well as symmetry; symmetry alone would establish real eigenvalues but not their interval location.
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