From the following:
$$\sum_0^\infty [ n(n-1)a_nx^{n-2} - n(n-1)a_n x^n -2na_nx^n + l(l+1)a_n x^n ] = 0$$ (a)
I'm trying to get to:
$$\sum_0^\infty [ (n+2)(n+1)a_{n+2} - [n(n+1) + l(l+1)]a_n]x^n = 0$$ (b)
Unfortunately, I'm not being very successful.
I can either get the first term ($(n+2)(n+1)a_{n+2} x^n$) by saying that $n=n+2$ or get the 2nd and 3rd terms $[n(n+1) + l(l+1)]a_nx^n$ just by simple calculations.
So, how do I get from a) to b)?
IMO this is not your problem, but the sign of the $l(l+1)$ term is positive in (a) but negative in (b). You can split and re-combine the sums with index shift because the $n(n-1)$ term is zero for $n=0,1.\;$ Here my manipulations $$\sum_0^\infty \Big( n(n-1)a_nx^{n-2} - n(n-1)a_n x^n -2na_nx^n + l(l+1)a_n x^n \Big)=$$ $$\sum_0^\infty n(n-1)a_nx^{n-2} + \sum_0^\infty \left(- n(n-1) -2n + l(l+1) \right)a_n x^n=$$ $$\sum_2^\infty n(n-1)a_nx^{n-2} + \sum_0^\infty \left(- n(n+1) + l(l+1) \right)a_n x^n=$$ $$\sum_0^\infty (n+2)(n+1)a_{n+2}x^n + \sum_0^\infty \left(- n(n+1) + l(l+1) \right)a_n x^n=$$ $$\sum_0^\infty \Big( (n+2)(n+1)a_{n+2} - (n(n+1) - l(l+1))a_n \Big) x^n$$