Need help with putting results into a power series $$a_0\bigg[1-\frac{\lambda}{2!}x^2-\frac{(4-\lambda)\lambda}{4!}x^4-\frac{(8-\lambda)(4-\lambda)\lambda}{6!}x^6-\cdots \bigg]$$
$$a_1\bigg[x+\frac{2-\lambda}{3!}x^3+\frac{(6-\lambda)(2-\lambda)}{5!}x^5+\cdots \bigg].$$
Hey guys, can someone help me put these into a power series with respect to $x$. $a_0$ and $a_1$ on the outside are just constants which you may ignore. They are answers I've received from solving a differential equation but I'm having trouble putting each one into a series.
 A: Both are power series expanded at $x_0=0$ since they have the form
\begin{align*}
  \sum_{n=0}^\infty \lambda_nx^n
  \end{align*}
In fact both series are Maclaurin series and we could look for an expression using sigma notation.
Here we take a look at the first series.

  
*
  
*Since the exponents are even, we have an even power series $A(x)$, i.e. $A(x)=A(-x)$.
  \begin{align*}
  A(x)=\sum_{n=0}^\infty b_{2n}x^{2n}
  \end{align*}
  
*There is a  factor $(2n)!$ in the denominator of each term. So, we have in fact an exponential power series
\begin{align*}
  \sum_{n=0}^\infty \lambda_{2n}\frac{x^{2n}}{(2n)!}
  \end{align*}
  
  
*
  
*There are $n$ factors in the numerator of $x^{2n}$ of the form
  $$\lambda(4-\lambda)\cdots(4(n-1)-\lambda)=-\prod_{j=0}^{n-1}\left(4j-\lambda\right)$$
  
*We assume only the first term  $1$ has positive sign and we obtain
  \begin{align*}
  A(x)=a_0\left(1-\sum_{n=1}^\infty\prod_{j=0}^{n-1}\left(4j-\lambda\right)\frac{x^{2n}}{(2n)!}\right)
  \end{align*}

With similar reasoning we can represent the second series in sigma notation.
A: Assuming this is a sum which goes up to $a_n[\cdots]$, the expression required is
$$\color{red}{\sum_{r=0}^na_r\sum_{j=0}^\infty \frac {(2(r+2j-2)-\lambda)!!!!\;\; x^{r+2j}}{(r+2j)!}}$$
where $(m-\lambda)!!!!$ is a specifically defined quadruple factorial, i.e. 
$$(m-\lambda)!!!!=(m-\lambda)(m-4-\lambda)(m-8-\lambda)\cdots ((m-1)\mod 4+1-\lambda)$$
and
$(m-\lambda)!!!!=1$ for $m<0$.

NB: For the coefficient of $a_0$, it's helpful to write $-\lambda$ as $(0-\lambda)$, in which case the expression shown in the original question becomes
$$a_0\bigg[1+\frac{(0-\lambda)}{2!}x^2+\frac{(4-\lambda)(0-\lambda)}{4!}x^4+\frac{(8-\lambda)(4-\lambda)(0-\lambda)}{6!}x^6+\cdots \bigg]$$
