Problems 7-3: 1acd, 2
Recursion formulas:
For 2a: $a_n = -\frac{1}{(n + \alpha - 2)(n + \alpha - 1)}a_{n-2}$, $n = 2, 3, 4, \ldots$
For 2b: $a_{n+1} = -\frac{2}{(n + \alpha + 1)(n + \alpha - 1)}a_n$, $n = 0, 1, 2, \ldots$