[9] \dfrac{(243)^{n/5} \times 3^{2n+1}}{9^{n}\times 3^{n-1}}=?

For Competitive Exams

$\dfrac{(243)^{n/5} \times 3^{2n+1}}{9^{n}\times 3^{n-1}}$=?

3ⁿ

Explanation:

Given Expression = $\dfrac{(243)^{n/5} \times 3^{2n+1}}{9^{n}\times 3^{n-1}}$

=$\dfrac{(3^{5})^{(n/5)} \times 3^{2n+1}}{(3^{2})^{n}\times 3^{n-1}}$

=$\dfrac{(3^{5\times(n/5)} \times 3^{2n+1})}{(3^{2n}\times 3^{n-1})}$

=$\dfrac{(3^{n} \times 3^{2n+1})}{(3^{2n}\times 3^{n-1})}$

=$\dfrac{3^{(n+2n+1)}}{3^{(2n+n-1)}}$

=$\dfrac{3^{(3n+1)}}{3^{(3n-1)}}$

= 3^{(3n + 1 - 3n + 1)} = 3² = 9.

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