44. The value of \( \frac{(243)^{\frac{n}{5}} \cdot 3^{2 n+1}}{9^{n} \cdot 3^{n-1}} \) is:

To simplify the expression \( \frac{(243)^{\frac{n}{5}} \cdot 3^{2 n+1}}{9^{n} \cdot 3^{n-1}} \), we first express 243 and 9 as powers of 3: - We know that \( 243 = 3^5 \), so \( (243)^{\frac{n}{5}} = (3^5)^{\frac{n}{5}} = 3^{5 \cdot \frac{n}{5}} = 3^n \). - We also know that \( 9 = 3^2 \), thus \( 9^{n} = (3^2)^n = 3^{2n} \). Now, we can rewrite the original expression: \[ \frac{(243)^{\frac{n}{5}} \cdot 3^{2 n + 1}}{9^{n} \cdot 3^{n - 1}} = \frac{3^n \cdot 3^{2n + 1}}{3^{2n} \cdot 3^{n - 1}}. \] Combining the exponents in the numerator, we have: \[ 3^n \cdot 3^{2n + 1} = 3^{n + 2n + 1} = 3^{3n + 1}. \] In the denominator, combining the exponents gives: \[ 3^{2n} \cdot 3^{n - 1} = 3^{2n + n - 1} = 3^{3n - 1}. \] Now our expression simplifies to: \[ \frac{3^{3n + 1}}{3^{3n - 1}}. \] Using the property of exponents, \( \frac{a^m}{a^n} = a^{m-n} \), we have: \[ 3^{(3n + 1) - (3n - 1)} = 3^{3n + 1 - 3n + 1} = 3^{2}. \] Thus, the value of the expression is: \[ \boxed{9}. \]