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 May 20, 2021
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$a_0 = 54 \div 3^3 = \frac{2}{3} = 2 \cdot 3^{-1}$

$a_1 = \frac{2}{3} \cdot 3 = 2 = 2 \cdot 3^0$

$a_2 = 2 \cdot 3 = 6 = 2 \cdot 3^1$

$a_3 = 6 \cdot 3 = 18 = 2 \cdot 3^2$

$a_4 = 18 \cdot 3 = 54 = 2 \cdot 3^3$

It should be clear by now that for $a_n$, it will be $\boxed{2 \cdot 3^{n-1}}.$

 May 20, 2021

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