The expression `2^(6n)-4^(2n)` where n is a natural number is always divisible by

Options

A. 15
B. 18
C. 36
D. 48

Answer

To solve the expression \(2^{6n} - 4^{2n}\) and determine what it is always divisible by, we can follow these steps:

### Step 1: Rewrite the expression
The expression given is:
\[ 2^{6n} - 4^{2n} \]
We know that \(4^{2n}\) can be rewritten in terms of base 2:
\[ 4^{2n} = (2^2)^{2n} = 2^{4n} \]
Thus, we can rewrite the expression as:
\[ 2^{6n} - 2^{4n} \]

### Step 2: Factor the expression
Now we can factor out the common term \(2^{4n}\):
\[ 2^{6n} - 2^{4n} = 2^{4n}(2^{2n} - 1) \]

### Step 3: Analyze the factors
Now we have:
\[ 2^{4n}(2^{2n} - 1) \]
The term \(2^{4n}\) is clearly divisible by \(2^{4n}\), which is a power of 2.

### Step 4: Determine divisibility of \(2^{2n} - 1\)
Next, we need to analyze \(2^{2n} - 1\). This expression is a difference of squares:
\[ 2^{2n} - 1 = (2^n - 1)(2^n + 1) \]
Both \(2^n - 1\) and \(2^n + 1\) are consecutive integers, meaning one of them is always even. Therefore, \(2^{2n} - 1\) is always divisible by 2.

### Step 5: Combine the factors
Since \(2^{4n}\) is divisible by \(2^{4n}\) and \(2^{2n} - 1\) is divisible by 2, we can conclude that:
\[ 2^{4n}(2^{2n} - 1) \]
is always divisible by \(2^{4n + 1}\).

### Conclusion
Thus, the expression \(2^{6n} - 4^{2n}\) is always divisible by \(2^{4n + 1}\).