Which is bigger: $A$ or $B$?

$\begin{aligned} A & =\dfrac{2+2^2+2^3+\dots+2^{3n-1}}{\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dots+\dfrac{1}{2^{3n-1}}} \\ \ \\ B & =\dfrac{3+3^2+3^3+\dots+3^{2n-1}}{\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dots+\dfrac{1}{3^{2n-1}}}\end{aligned}$

**Note:** $n \in \mathbb N$.

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