Simplify: `frac(16 xx 2^(n+1)-4 xx 2^(n))(16 xx 2^(n+2)-2 xx 2^(n+2)).`

Answer

To simplify the expression \(\frac{16 \cdot 2^{(n+1)} - 4 \cdot 2^n}{16 \cdot 2^{(n+2)} - 2 \cdot 2^{(n+2)}}\), we can follow these steps:

### Step 1: Rewrite the expression
Start with the original expression:
\[ \frac{16 \cdot 2^{(n+1)} - 4 \cdot 2^n}{16 \cdot 2^{(n+2)} - 2 \cdot 2^{(n+2)}} \]

### Step 2: Factor out common terms
In the numerator, we can factor out \(4 \cdot 2^n\):
\[ = \frac{4 \cdot 2^n (4 \cdot 2 - 1)}{16 \cdot 2^{(n+2)} - 2 \cdot 2^{(n+2)}} \]
This simplifies to:
\[ = \frac{4 \cdot 2^n (8 - 1)}{16 \cdot 2^{(n+2)} - 2 \cdot 2^{(n+2)}} \]

### Step 3: Simplify the numerator
Now simplify the numerator:
\[ = \frac{4 \cdot 2^n \cdot 7}{16 \cdot 2^{(n+2)} - 2 \cdot 2^{(n+2)}} \]

### Step 4: Simplify the denominator
In the denominator, factor out \(2^{(n+2)}\):
\[ = \frac{4 \cdot 2^n \cdot 7}{(16 - 2) \cdot 2^{(n+2)}} \]
This simplifies to:
\[ = \frac{4 \cdot 2^n \cdot 7}{14 \cdot 2^{(n+2)}} \]

### Step 5: Cancel common factors
Now, we can cancel \(2^n\) from the numerator and denominator:
\[ = \frac{4 \cdot 7}{14 \cdot 2^2} \]

### Step 6: Simplify further
Now simplify the fraction:
\[ = \frac{28}{56} = \frac{1}{2} \]

### Final Answer
Thus, the simplified expression is:
\[ \frac{1}{2} \]

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