\( \frac{2^{n+1} 9^{n-2}}{6^{n-1}-3^{n+1}} \). Writo your answer as a fraction in its simpleat form.

To simplify the expression \( \frac{2^{n+1} 9^{n-2}}{6^{n-1}-3^{n+1}} \), we start by rewriting the terms: 1. Rewrite \( 9^{n-2} \) as \( (3^2)^{n-2} = 3^{2(n-2)} = 3^{2n-4} \). 2. Rewrite \( 6^{n-1} \) as \( (2 \cdot 3)^{n-1} = 2^{n-1} \cdot 3^{n-1} \). 3. Then, \( 6^{n-1} - 3^{n+1} = 2^{n-1} \cdot 3^{n-1} - 3^{n+1} \). Now, substituting into the expression: \[ = \frac{2^{n+1} \cdot 3^{2n-4}}{2^{n-1} \cdot 3^{n-1} - 3^{n+1}} \] The denominator simplifies further: \[ 2^{n-1} \cdot 3^{n-1} - 3^{n+1} = 3^{n-1}(2^{n-1} - 3^2) = 3^{n-1}(2^{n-1} - 9) \] Now we can pour everything back together: \[ = \frac{2^{n+1} \cdot 3^{2n-4}}{3^{n-1}(2^{n-1} - 9)} \] This gives us: \[ = \frac{2^{n+1}}{3^{n-1}} \cdot \frac{3^{2n-4}}{2^{n-1}-9} \] \[ = \frac{2^{n+1} \cdot 3^{2n-4}}{3^{n-1} (2^{n-1} - 9)} \] Finally simplifying further: \[ = \frac{2^{n+1}}{3^{n-1}} \cdot \frac{3^{n-1} \cdot 3^{n-2}}{(2^{n-1}-9)} \] \[ = \frac{2^{n+1} \cdot 3^{n-2}}{2^{n-1} - 9} \] This final form clearly shows the components in fraction, ensuring it's simplified to: \[ \frac{2^{n+1} \cdot 3^{n-2}}{2^{n-1} - 9} \] That’s your simplified fraction!