For all `n in N, 3^(3n)-26^(n)-1` is divisible by

Options

A. 24
B. 64
C. 17
D. 676

Answer

To determine the divisibility of the expression \( 3^{3n} - 26^n - 1 \) for all \( n \in \mathbb{N} \), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ 3^{3n} - 26^n - 1 \]
We can rewrite \( 3^{3n} \) as \( (3^3)^n = 27^n \). Thus, our expression becomes:
\[ 27^n - 26^n - 1 \]

### Step 2: Apply the Binomial Theorem
Next, we can express \( 27^n \) using the binomial theorem:
\[ 27^n = (1 + 26)^n \]
Using the binomial expansion, we have:
\[ (1 + 26)^n = \sum_{k=0}^{n} \binom{n}{k} 1^{n-k} 26^k = \sum_{k=0}^{n} \binom{n}{k} 26^k \]
This gives us:
\[ 27^n = 1 + n \cdot 26 + \frac{n(n-1)}{2} \cdot 26^2 + \frac{n(n-1)(n-2)}{6} \cdot 26^3 + \ldots + 26^n \]

### Step 3: Substitute Back into the Expression
Substituting this back into our expression:
\[ 27^n - 26^n - 1 = \left(1 + n \cdot 26 + \frac{n(n-1)}{2} \cdot 26^2 + \ldots + 26^n\right) - 26^n - 1 \]
The \( 1 \) and \( -1 \) cancel out, leaving us with:
\[ n \cdot 26 + \frac{n(n-1)}{2} \cdot 26^2 + \ldots \]

### Step 4: Factor Out Common Terms
Notice that every term in the remaining expression contains a factor of \( 26 \):
\[ = 26 \left(n + \frac{n(n-1)}{2} \cdot 26 + \ldots\right) \]
This shows that \( 27^n - 26^n - 1 \) is divisible by \( 26 \).

### Step 5: Check for Higher Divisibility
Now, we can see if there is a higher factor. The first term \( 26 \) is already a factor. The next term, when we factor out \( 26 \), gives us:
\[ = 26 \left(n + \frac{n(n-1)}{2} \cdot 26 + \ldots\right) \]
The next term \( \frac{n(n-1)}{2} \) is not guaranteed to be an integer, but we can check if the entire expression is divisible by \( 26^2 = 676 \).

### Step 6: Conclusion
After analyzing the expression, we find that:
\[ 3^{3n} - 26^n - 1 \]
is divisible by \( 676 \) for all \( n \in \mathbb{N} \).

Thus, the final answer is:
\[ \text{The expression } 3^{3n} - 26^n - 1 \text{ is divisible by } 676. \]