The value of `(6^(n+3) - 32.6^(n+1))/(6^(n+2) - 2.6^(n+1))` is equal to

Options

A. 36
B. `1/6`
C. 2
D. 1

Answer

To solve the expression \(\frac{6^{n+3} - 32 \cdot 6^{n+1}}{6^{n+2} - 2 \cdot 6^{n+1}}\), we will simplify both the numerator and the denominator step by step.

### Step 1: Factor out common terms from the numerator
The numerator is \(6^{n+3} - 32 \cdot 6^{n+1}\). We can factor out \(6^{n+1}\):
\[ 6^{n+3} - 32 \cdot 6^{n+1} = 6^{n+1}(6^2 - 32) = 6^{n+1}(36 - 32) = 6^{n+1} \cdot 4 \]

### Step 2: Factor out common terms from the denominator
The denominator is \(6^{n+2} - 2 \cdot 6^{n+1}\). We can factor out \(6^{n+1}\):
\[ 6^{n+2} - 2 \cdot 6^{n+1} = 6^{n+1}(6 - 2) = 6^{n+1} \cdot 4 \]

### Step 3: Substitute the factored forms back into the expression
Now we can substitute the factored forms of the numerator and denominator back into the expression:
\[ \frac{6^{n+1} \cdot 4}{6^{n+1} \cdot 4} \]

### Step 4: Simplify the expression
Since both the numerator and the denominator are the same, we can simplify:
\[ \frac{6^{n+1} \cdot 4}{6^{n+1} \cdot 4} = 1 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{1} \]