`9^(n)-8^(n)-1` is divisible by 64 is

Options

A. Always true
B. Always false
C. Always true for rational values of n
D. Always true for irrational values of n.

Answer

To determine when \(9^n - 8^n - 1\) is divisible by 64, we can analyze the expression step by step.

### Step 1: Rewrite the Expression
We start with the expression:
\[ 9^n - 8^n - 1 \]
We can rewrite \(9^n\) as \((8 + 1)^n\):
\[ (8 + 1)^n - 8^n - 1 \]

### Step 2: Apply the Binomial Theorem
Using the Binomial Theorem, we 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:
\[ 1 + n \cdot 8 + \frac{n(n-1)}{2} \cdot 8^2 + \ldots + 8^n \]

### Step 3: Substitute Back into the Expression
Substituting this back into our expression:
\[ (1 + n \cdot 8 + \frac{n(n-1)}{2} \cdot 8^2 + \ldots + 8^n) - 8^n - 1 \]
This simplifies to:
\[ n \cdot 8 + \frac{n(n-1)}{2} \cdot 8^2 + \ldots + 8^{n-1} \]

### Step 4: Factor Out 8
Notice that every term in the expression contains a factor of 8:
\[ 8 \left( n + \frac{n(n-1)}{2} \cdot 8 + \ldots + 8^{n-2} \right) \]
This shows that \(9^n - 8^n - 1\) is divisible by 8.

### Step 5: Check for Divisibility by 64
Now we need to check if the expression inside the parentheses is divisible by 8:
\[ n + \frac{n(n-1)}{2} \cdot 8 + \ldots + 8^{n-2} \]
For \(9^n - 8^n - 1\) to be divisible by 64, we need:
\[ n + \frac{n(n-1)}{2} \cdot 8 + \ldots + 8^{n-2} \equiv 0 \mod 8 \]

### Step 6: Analyze the Conditions
1. If \(n\) is even, say \(n = 2k\), then \(n \equiv 0 \mod 8\).
2. If \(n\) is odd, say \(n = 2k + 1\), then \(n \equiv 1 \mod 8\).

Thus, the expression \(9^n - 8^n - 1\) is divisible by 64 when \(n\) is even.

### Final Conclusion
The expression \(9^n - 8^n - 1\) is divisible by 64 if \(n\) is even.

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