Prove by induction that $a_{2m}=\frac{(0-b)(1+b)(2-b)(3+b)....(2m-1+b)}{(2m)!}a_0$

Question

This is part of another question with a given ODE: $(1-x^2)y''-2xy'+b(b+1)y=0$.

Before my question I was asked to prove that if $y(x)=\sum_{n=0}^\infty a_nx^n$ is analytic solution of the ODE around $x_0=0$ then:

$a_{n+2}=\frac{(n-b)(n+1+b)}{(n+2)(n+1)}a_n$

Now, im supposed to prove by induction using this. I got stuck because I can't figure out what to do with $a_0$

Answer 1

Set $n=2m-2,\forall m=1,2,...$ at the given anadromic formula that describes $(a_n)$ and you get $$a_{2m}=\dfrac{(2m-p)(2m-1+b)}{2m\cdot(2m-1)},\ m=1,2,... $$ Now, if you write previous equations for all 1,2,...,m and then you multiply all of them you get the analytic form that describes the subsequence $(a_{2m})$ of $(a_n)$.