Start by expanding \(\frac{3(2^{n+1}) - 4(2^{n-1})}{2^{n+1} - 2^n}\): \(\frac{3 \times 2^n \times 2^1 - 2^2 \times 2^n \times 2^{-1}}{2^n \times 2 - 2^n}\) NUMERATOR : 2\(^n\) ( 3\(^1\) X 2\(^1\) - 2\(^2\) X 2\(^{-1}\) ) --> 2\(^n\) ( 3 X 2 — 4 X \(\frac{1}{2}\) ) --> 2\(^n\) ( 6 - 2 ) --> 2\(^n\) (4) DENOMINATOR : 2\(^n\) ( 2\(^1\) - 1 ) --> 2\(^n\) ( 2 - 1) --> 2\(^n\) => \(\frac{2^n ( 4)}{2^n}\) = 4