Solution
= Solution
Differentiate the <power series> twice and align the <coefficient> of $\xi^k$. The equation becomes
$$
(k+2)(k+1)a_{k+2}+(\epsilon-1-2k)a_k=0\qquad(k\geq0).
$$
Thus
$$
\boxed{a_{k+2}=\frac{2k+1-\epsilon}{(k+2)(k+1)}a_k.}
$$
Even <coefficients> are generated from $a_0$, and odd <coefficients> from $a_1$; the two <parity> <sequences> are <independent>.