`frac(3^(n+2) - 3^(n+1)) (3^(n+4) + 3^(n+1))` is equal to

Options

A. `frac(1)(5)`
B. `frac(1)(10)`
C. `frac(1)(14)`
D. `frac(1)(28)`

Answer

To solve the expression \(\frac{3^{n+2} - 3^{n+1}}{3^{n+4} + 3^{n+1}}\), we will simplify it step by step.

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

### Step 2: Factor out common terms in the denominator
The denominator is \(3^{n+4} + 3^{n+1}\). We can factor out \(3^{n+1}\):
\[ 3^{n+4} + 3^{n+1} = 3^{n+1}(3^3 + 1) = 3^{n+1}(27 + 1) = 3^{n+1} \cdot 28 \]

### Step 3: Substitute the factored forms back into the expression
Now we substitute the factored forms back into the original expression:
\[ \frac{3^{n+1} \cdot 2}{3^{n+1} \cdot 28} \]

### Step 4: Cancel out the common terms
We can cancel \(3^{n+1}\) from the numerator and the denominator:
\[ \frac{2}{28} \]

### Step 5: Simplify the fraction
Now we simplify \(\frac{2}{28}\):
\[ \frac{2}{28} = \frac{1}{14} \]

Thus, the final answer is:
\[ \frac{1}{14} \]