



 
Hi Chelsea, I also get $ \frac{4}{17}.$ $243 = 3^5$ and $81 = 3^4$ so \[\frac{81^{3n + 2}}{243^{n}}\] becomes \[\left( 3^4 \right)^{3n + 2} \times \left( 3^5 \right)^n = 3^4\] Is that what you did?




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