Question

(series solution differential equations) if a0=0 why is a2k=0?

Answer 1

because it looks like in the solution it is revealed that:
\[a_n = (blah)a_{n-2}\]
where blah is some constant. Since a_0 is 0, that means:
\[a_2 = (blah)(0) = 0, a_4 = (blah)(0) = 0, \ldots \]
Every even term is going to be 0, so they just say it like this:
\[a_{2k} = 0, k = 1,2,3,\ldots\]