I need help on deriving the general formula for the Taylor series of $frac{2x}{e^{2x}-1}$

It is known that the series for $dfrac{x}{e^{x}-1}$ is: $$dfrac{x}{e^{x}-1}=sum_{n=0}^{+infty}dfrac{B_nx^{n}}{n!}$$

So the question is whether the series for $frac{2x}{e^{2x}-1}$ is:

$$frac{2x}{e^{2x}-1}=sum_{n=0}^{+infty}dfrac{B_n(2x)^{n}}{n!}tag{1}$$

or

$$frac{2x}{e^{2x}-1}=2left(dfrac{x}{e^{2x}-1}right)=2left(sum_{n=0}^{+infty}dfrac{B_n(2x)^{n}}{n!}right)tag{2}$$

So which series is the correct representation for $frac{2x}{e^{2x}-1}$? $(1)$ or $(2)$

I suspect that it must be $(2)$, because if it is $(1)$, how does one represent the series for $dfrac{3x}{e^{2x}-1}$