If `n in N`, then `3^(2n)+7` is divisible by

Options

A. 3
B. 8
C. 9
D. 11

Answer

To solve the problem, we need to show that \(3^{2n} + 7\) is divisible by a certain number for all \(n \in \mathbb{N}\). Let's analyze the expression step by step.

### Step 1: Rewrite the expression
We start with the expression \(3^{2n} + 7\). Notice that \(3^{2n}\) can be rewritten as \((3^2)^n = 9^n\).

\[ 3^{2n} + 7 = 9^n + 7 \]

### Step 2: Factor the expression
Next, we can express \(9^n + 7\) in a way that allows us to analyze its divisibility. We can rewrite \(9^n\) as \(8 + 1\):

\[ 9^n + 7 = (8 + 1)^n + 7 \]

### Step 3: Expand using the Binomial Theorem
Using the Binomial Theorem, we can expand \((8 + 1)^n\):

\[ (8 + 1)^n = \sum_{k=0}^{n} \binom{n}{k} 8^k \cdot 1^{n-k} = \binom{n}{0} 8^0 + \binom{n}{1} 8^1 + \binom{n}{2} 8^2 + \ldots + \binom{n}{n} 8^n \]

This gives us:

\[ (8 + 1)^n = 1 + 8n + \frac{n(n-1)}{2} \cdot 8^2 + \ldots + 8^n \]

### Step 4: Combine with the constant
Now, we add 7 to the expanded expression:

\[ (8 + 1)^n + 7 = 1 + 8n + \frac{n(n-1)}{2} \cdot 8^2 + \ldots + 8^n + 7 \]

This simplifies to:

\[ = 8n + 8 + \frac{n(n-1)}{2} \cdot 8^2 + \ldots + 8^n \]

### Step 5: Factor out 8
We can factor out 8 from the entire expression:

\[ = 8 \left( n + 1 + \frac{n(n-1)}{16} + \ldots + \frac{8^{n-1}}{8} \right) \]

### Conclusion
Since the entire expression can be factored out by 8, we conclude that:

\[ 3^{2n} + 7 \text{ is divisible by } 8 \text{ for all } n \in \mathbb{N}. \]

### Final Answer
Thus, the answer is that \(3^{2n} + 7\) is divisible by **8**.

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