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