cumulants of non-central $\chi^2$ distribution

Question

Cumulant generating function is defined by logarithm of moment generating function.

$$K_X(t)=\log M_X(t)$$

Let $X$ be a non-central $\chi^2$ variate with parameters degrees of freedom, $n$ and non-centrality parameter $\lambda$.

Moment generating function of $X$ is:

$$ M_X(t)=(1-2t)^{-\frac{n}{2}}\exp\left(\frac{\lambda t}{1-2t}\right)$$ $$ K_X(t)=\log M_X(t)=\log \left[(1-2t)^{-\frac{n}{2}}\exp\left(\frac{\lambda t}{1-2t}\right)\right]$$ $$ K_X(t)= -\frac{n}{2} \log (1-2t)+\left(\frac{\lambda t}{1-2t}\right)$$ $$ K_X(t)= -\frac{n}{2} \left[-2t-\frac{(2t)^2}{2}-\frac{(2t)^3}{3}-\frac{(2t)^4}{4}-\ldots\right]+\left(\frac{\lambda t}{1-2t}\right) \tag{A}$$

$r^{th}$ cumulant is the coefficient of $\frac{t^r}{r!}$ in $K_X(t)$.

But i couldn't separate $\frac{t^r}{r!}$ in $K_X(t)$ because of the denominator $(1-2t)$. Hence couldn't able to compute the cumulants to know the mean,variance, 3rd&4th central moment of the distribution.

Can you please rearrange the equation $(A)$ as $\frac{t^r}{r!}$ so that the four cumulants can be obtained easily.

Answer 1

$$ \begin{align} \frac{\lambda t}{1-2t} & = \lambda t\left(1+2t+4t^2-8t^3+\cdots+(2t)^k+\cdots\right) \\[10pt] & = \lambda t + 2\lambda t^2 + 4\lambda t^3 + 8\lambda t^4 + \cdots + \lambda t(-2t)^{k}+\cdots \end{align} $$

(It's a geometric series.)