`2^(3n)-7n-1` is divisible by

Options

A. 64
B. 36
C. 49
D. 25

Answer

To determine the divisibility of the expression \(2^{3n} - 7n - 1\), we can analyze it step by step.

### Step 1: Rewrite the Expression
We start with the expression:
\[ 2^{3n} - 7n - 1 \]
We can express \(2^{3n}\) as \((2^3)^n = 8^n\). Thus, we rewrite the expression:
\[ 8^n - 7n - 1 \]

### Step 2: Apply the Binomial Theorem
Next, we can use the Binomial Theorem to expand \((7 + 1)^n\):
\[ (7 + 1)^n = \sum_{k=0}^{n} \binom{n}{k} 7^{n-k} \cdot 1^k \]
This gives us:
\[ 7^n + \binom{n}{1} 7^{n-1} + \binom{n}{2} 7^{n-2} + \ldots + 1 \]
Now, we can express \(8^n\) in terms of this expansion.

### Step 3: Compare Terms
We can compare the terms of \(8^n\) and the expansion of \((7 + 1)^n\):
\[ 8^n = (7 + 1)^n = 7^n + \binom{n}{1} 7^{n-1} + \binom{n}{2} 7^{n-2} + \ldots + 1 \]
Subtracting \(7n + 1\) from both sides gives:
\[ 8^n - 7n - 1 = 7^n + \binom{n}{1} 7^{n-1} + \binom{n}{2} 7^{n-2} + \ldots + 1 - 7n - 1 \]

### Step 4: Simplify the Expression
Now, we simplify the expression:
\[ = 7^n + \binom{n}{1} 7^{n-1} + \binom{n}{2} 7^{n-2} + \ldots + 1 - 7n - 1 \]
Notice that the terms involving \(7n\) will cancel out with the corresponding terms in the expansion.

### Step 5: Identify Divisibility
After simplification, we find that the remaining terms are multiples of \(49\) (since \(7^2 = 49\)). Thus, we conclude that:
\[ 2^{3n} - 7n - 1 \text{ is divisible by } 49 \]

### Final Answer
Thus, the expression \(2^{3n} - 7n - 1\) is divisible by \(49\).

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