Solving the recurrence $a_{k+2} = \frac {k - 4}{(k+1)(k+2)}a_k$? I am solving the ODE $y'' - xy' + 4y = 0$ via power series, which leads me to the following recurrence:
$$a_{k+2} = \frac {k - 4}{(k+1)(k+2)}a_k \tag{1}$$
where $k \ge 1$ and $a_0, a_1$ are given. I would like to ask a question about one of the steps in my solution ($\color{red}{\text{in red below}}$), and whether there is a better solution.
My Solution:
With $k = 2$ we see that $a_4 = 0$, and as a result $a_n = 0$ for all even $n \ge 4$. So we concentrate on $a_n$ with $n$ odd.
Define $b_k = a_{2k + 1}$ for $k \ge 0$. Letting $k = 2k-1$ in $(1)$,
$$b_k = \frac {2k - 5}{2k(2k+1)}b_{k-1}.$$
Now
$$
\begin{align*}
b_k &=  \frac {2k - 5}{2k(2k+1)}b_{k-1} \\
&= \frac {2k - 5}{2k(2k+1)} \cdot \frac {2k - 7}{2(k-1)(2k-1)}b_{k-2}\\
 & \ \ \vdots\\
&= \frac {2k - 5}{2k(2k+1)} \cdot \frac {2k - 7}{2(k-1)(2k-1)} \cdots \frac {-3}{2(1)(1)}b_0
\end{align*}.
$$
$\color{red}{\text{(trouble above)}}$
After cancelling, we have
$$b_k = \frac {3}{2^k \cdot k! \cdot (2k+1)(2k-1)(2k-3)}b_0$$
and we are essentially done.
Trouble: I am worried about the part where I used the vertical dots. I always have difficulty with figuring out what the final coefficient is once we get to $b_0$. This is how I currently think about it:
Consider $b_k =  \frac {2k - 5}{2k(2k+1)}b_{k-1}$. To get to $b_0$ on the RHS, we must subtract $1$ a total number of $k-1$ times. Every time we subtract a $1$, we subtract a $2$ from the $(2k-5)$ term, so the (numerator of the) final term should be $2k-5 - 2(k-1) = -3$. Similarly for the $(2k)$ and $(2k+1)$ terms in the denominator.
In which way do you figure out the last term? (I think my way is not the best)

Introduce additional cancelling terms
$$
b_k =  \frac {2k - 5}{2k(2k+1)}b_{k-1} 
=\frac {(2k-1)(2k-3)(2k - 5)}{2k(2k+1)(2k-1)(2k-3)}b_{k-1}  
$$
to see that
\begin{align}
2^kk!(2k+1)(2k-1)(2k-3)b_k&=2^{k-1}(k-1)!(2k-1)(2k-3)(2k - 5)b_{k-1}\\
=\dots &=2^00!(2·0+1)(2·0-1)(2·0-3)b_0=3b_0
\end{align}
The first equality shows that the expression $2^kk!(2k+1)(2k-1)(2k-3)b_k$ is constant in $k$, and then inserting $k=0$ gives the value of that constant in terms of $b_0$.
