Question 120
For each nonnegative integer $n$, polynomial $K_n(x_1, x_2, ..., x_n)$ is defined recursively as follows,
$$
\begin{array}{l}
K_0 = 1 \\
K_1(x_1) = x_1 \\
K_n(x_1, \dots, x_n) = x_n K_{n-1}(x_1, \dots, x_{n-1}) + (x_n^2 + x_{n-1}^2) K_{n-2}(x_1, \dots, x_{n-2}).
\end{array}
$$
Prove that $K_n(x_1, x_2, ..., x_{n-1}, x_n) = K_n(x_n, x_{n-1}, ..., x_2, x_1)$.
$$
\begin{array}{l}
K_0 = 1 \\
K_1(x_1) = x_1 \\
K_n(x_1, \dots, x_n) = x_n K_{n-1}(x_1, \dots, x_{n-1}) + (x_n^2 + x_{n-1}^2) K_{n-2}(x_1, \dots, x_{n-2}).
\end{array}
$$
Prove that $K_n(x_1, x_2, ..., x_{n-1}, x_n) = K_n(x_n, x_{n-1}, ..., x_2, x_1)$.