1) \( \frac{3^{n+4} \cdot 6 \cdot 3^{n+1}}{7 \cdot 3^{n+2}} \)

First, let's simplify the expression step-by-step. We start with: \[ \frac{3^{n+4} \cdot 6 \cdot 3^{n+1}}{7 \cdot 3^{n+2}} \] In the numerator, we can combine the powers of \(3\): \[ 3^{n+4} \cdot 3^{n+1} = 3^{(n+4) + (n+1)} = 3^{2n+5} \] So, the expression now looks like: \[ \frac{6 \cdot 3^{2n+5}}{7 \cdot 3^{n+2}} \] Next, we can simplify this even further by dividing the powers of \(3\): \[ \frac{3^{2n+5}}{3^{n+2}} = 3^{(2n+5) - (n+2)} = 3^{n+3} \] Now, substituting this back into the fraction gives us: \[ \frac{6 \cdot 3^{n+3}}{7} \] Thus, the final simplified expression is: \[ \frac{6 \cdot 3^{n+3}}{7} \]