\( \frac{\text { 2.L) } 5 \cdot 2^{n+3}-3 \cdot 2^{n+2}}{2^{n+1}} \)

Let's simplify the expression: \[ \frac{5 \cdot 2^{n+3} - 3 \cdot 2^{n+2}}{2^{n+1}} \] **Step 1: Factor Out Common Terms in the Numerator** First, recognize that \(2^{n+3}\) and \(2^{n+2}\) can both be expressed in terms of \(2^{n+1}\): \[ 2^{n+3} = 2^{n+1} \cdot 2^2 = 2^{n+1} \cdot 4 \] \[ 2^{n+2} = 2^{n+1} \cdot 2 = 2^{n+1} \cdot 2 \] Now, substitute these back into the numerator: \[ 5 \cdot 2^{n+3} = 5 \cdot 2^{n+1} \cdot 4 = 20 \cdot 2^{n+1} \] \[ 3 \cdot 2^{n+2} = 3 \cdot 2^{n+1} \cdot 2 = 6 \cdot 2^{n+1} \] So, the numerator becomes: \[ 20 \cdot 2^{n+1} - 6 \cdot 2^{n+1} = 14 \cdot 2^{n+1} \] **Step 2: Simplify the Fraction** Now, the expression looks like this: \[ \frac{14 \cdot 2^{n+1}}{2^{n+1}} \] Since \(2^{n+1}\) appears in both the numerator and the denominator, they cancel each other out: \[ \frac{14 \cdot \cancel{2^{n+1}}}{\cancel{2^{n+1}}} = 14 \] **Final Answer:** The simplified expression is **14**.