For each `n in N, 3^(2n)-1` is divisible by

Options

A. (a) 8
B. (b) 16
C. (c) 32
D. (d) 10

Answer

To solve the problem of determining by which number the expression \(3^{2n} - 1\) is divisible for each \(n \in \mathbb{N}\), we can follow these steps:

### Step 1: Define the expression
Let \(P(n) = 3^{2n} - 1\).

### Step 2: Evaluate for the first few natural numbers
We will evaluate \(P(n)\) for the first few natural numbers to find a pattern.

1. For \(n = 1\):
\[ P(1) = 3^{2 \cdot 1} - 1 = 3^2 - 1 = 9 - 1 = 8 \]

2. For \(n = 2\):
\[ P(2) = 3^{2 \cdot 2} - 1 = 3^4 - 1 = 81 - 1 = 80 \]

3. For \(n = 3\):
\[ P(3) = 3^{2 \cdot 3} - 1 = 3^6 - 1 = 729 - 1 = 728 \]

### Step 3: Check divisibility
Now, we will check the divisibility of the results obtained:

- \(P(1) = 8\) is divisible by \(8\).
- \(P(2) = 80\) is divisible by \(8\).
- \(P(3) = 728\) is divisible by \(8\) (since \(728 \div 8 = 91\)).

### Step 4: Generalize the result
We observe that:
- \(P(n) = 3^{2n} - 1\) can be factored using the difference of squares:
\[ P(n) = (3^n - 1)(3^n + 1) \]

For \(n \geq 1\):
- \(3^n - 1\) is always even (since \(3^n\) is odd).
- \(3^n + 1\) is also even.

Thus, the product \((3^n - 1)(3^n + 1)\) is divisible by \(4\).

### Step 5: Check higher powers of \(2\)
To check if \(P(n)\) is divisible by \(8\), we can analyze the terms:
- For \(n = 1\), \(P(1) = 8\).
- For \(n = 2\), \(P(2) = 80\) which is also divisible by \(8\).
- For \(n = 3\), \(P(3) = 728\) which is divisible by \(8\).

### Conclusion
From our evaluations, we can conclude that \(3^{2n} - 1\) is divisible by \(8\) for all \(n \in \mathbb{N}\).

Thus, the answer is:
\[ \text{Option A: } 8 \]