A) 1
B) 9
C) 3
D) \[{{3}^{n}}\]
Correct Answer: B
Solution :
\[\frac{{{(243)}^{\frac{n}{5}}}\cdot {{3}^{2n+1}}}{{{9}^{n}}\cdot {{3}^{n-1}}}=\frac{{{({{3}^{5}})}^{\frac{n}{5}}}\cdot {{3}^{2n+1}}}{{{3}^{2n}}\cdot {{3}^{n-1}}}\] \[=\frac{{{3}^{n}}\cdot {{3}^{2n+1}}}{{{3}^{2n}}\cdot {{3}^{n-1}}}=\frac{{{3}^{3n+1}}}{{{3}^{3n-1}}}=\frac{{{3}^{2}}\cdot {{3}^{3n-1}}}{{{3}^{3n-1}}}\] \[={{3}^{2}}=9\]You need to login to perform this action.
You will be redirected in
3 sec