`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 whether \( 9^n - 8^n - 1 \) is divisible by 64, we will analyze the expression step by step.

### Step 1: Rewrite the Expression
We start with the expression:
\[ 9^n - 8^n - 1 \]

### Step 2: Use the Binomial Theorem
We can express \( 9^n \) using the binomial theorem. Note that \( 9 = 8 + 1 \), so we can expand \( (8 + 1)^n \):
\[ 9^n = (8 + 1)^n = \sum_{k=0}^{n} \binom{n}{k} 8^k 1^{n-k} = \binom{n}{0} 8^0 + \binom{n}{1} 8^1 + \binom{n}{2} 8^2 + \ldots + \binom{n}{n} 8^n \]
This simplifies to:
\[ 9^n = 1 + n \cdot 8 + \frac{n(n-1)}{2} \cdot 8^2 + \ldots + 8^n \]

### Step 3: Substitute Back into the Expression
Now substituting this back into our original expression:
\[ 9^n - 8^n - 1 = \left(1 + n \cdot 8 + \frac{n(n-1)}{2} \cdot 8^2 + \ldots + 8^n\right) - 8^n - 1 \]
The \( 1 \) and \( -1 \) cancel out, and we are left with:
\[ n \cdot 8 + \frac{n(n-1)}{2} \cdot 8^2 + \ldots \]

### Step 4: Factor Out 8
We can factor out \( 8 \) from the remaining terms:
\[ 9^n - 8^n - 1 = 8 \left(n + \frac{n(n-1)}{2} \cdot 8 + \ldots\right) \]
This shows that \( 9^n - 8^n - 1 \) is divisible by \( 8 \).

### Step 5: Check for Divisibility by 64
To check if it is divisible by \( 64 \), we need to analyze the expression further. We will consider the expression modulo \( 64 \).

#### Case Analysis:
1. **For \( n = 1 \)**:
\[ 9^1 - 8^1 - 1 = 9 - 8 - 1 = 0 \quad (\text{divisible by } 64) \]

2. **For \( n = 2 \)**:
\[ 9^2 - 8^2 - 1 = 81 - 64 - 1 = 16 \quad (\text{not divisible by } 64) \]

3. **For \( n = 3 \)**:
\[ 9^3 - 8^3 - 1 = 729 - 512 - 1 = 216 \quad (\text{not divisible by } 64) \]

4. **For \( n = 4 \)**:
\[ 9^4 - 8^4 - 1 = 6561 - 4096 - 1 = 2464 \quad (\text{divisible by } 64) \]

### Conclusion
From the analysis, we see that \( 9^n - 8^n - 1 \) is not consistently divisible by \( 64 \) for all \( n \). It is divisible for certain values of \( n \) (like \( n = 1 \) and \( n = 4 \)) but not for others (like \( n = 2 \) and \( n = 3 \)). Therefore, we conclude that \( 9^n - 8^n - 1 \) is not universally divisible by \( 64 \).