Step-by-step Solution:

Correct Answer: C ($A \subset B$)

To determine the relationship between sets A and B, we can analyze their elements and then prove the relationship mathematically.

Step 1: Analyze the Elements of Set A and Set B

Set A is defined as $A = \{5^n - 4n - 1 : n \in \mathbb{N}\}$. Let's find the first few elements:

• For $n=1$: $5^1 - 4(1) - 1 = 5 - 4 - 1 = 0$
• For $n=2$: $5^2 - 4(2) - 1 = 25 - 8 - 1 = 16$
• For $n=3$: $5^3 - 4(3) - 1 = 125 - 12 - 1 = 112$

So, $A = \{0, 16, 112, ...\}$.

Set B is defined as $B = \{16(n-1) : n \in \mathbb{N}\}$. Let's find the first few elements:

• For $n=1$: $16(1-1) = 0$
• For $n=2$: $16(2-1) = 16$
• For $n=3$: $16(3-1) = 32$
• For $n=4$: $16(4-1) = 48$

So, $B = \{0, 16, 32, 48, ...\}$. This is the set of all non-negative multiples of 16.

Step 2: Compare the Sets and Form a Hypothesis

By comparing the elements, we can see that the first few elements of A (0, 16, 112) are all multiples of 16 and are therefore also in B. However, B contains elements like 32 and 48 that are not in A. This suggests that A is a subset of B ($A \subset B$).

Step 3: Prove that $A \subset B$

To prove that A is a subset of B, we must show that every element of A is a multiple of 16. We can use the Binomial Theorem to prove that the expression $5^n - 4n - 1$ is always divisible by 16 for any natural number $n$.

Let's rewrite $5^n$ as $(1+4)^n$ and expand it:

$(1+4)^n = \binom{n}{0}1^n + \binom{n}{1}1^{n-1}(4)^1 + \binom{n}{2}1^{n-2}(4)^2 + \binom{n}{3}1^{n-3}(4)^3 + \dots$

$5^n = 1 + n(4) + \frac{n(n-1)}{2}(16) + \frac{n(n-1)(n-2)}{6}(64) + \dots$

$5^n = 1 + 4n + 16 \left( \frac{n(n-1)}{2} + \dots \right)$

Now, substitute this back into the expression for an element of A:

$5^n - 4n - 1 = \left(1 + 4n + 16 \left( \frac{n(n-1)}{2} + \dots \right)\right) - 4n - 1$

$5^n - 4n - 1 = 16 \left( \frac{n(n-1)}{2} + \dots \right)$

Since all the remaining terms in the expansion will contain a factor of $16$ or higher powers of 4, the entire expression is a multiple of 16. Therefore, every element of set A is a multiple of 16, which means every element of A is also in set B.

Conclusion

Since every element of A is in B, but not every element of B is in A, the correct relationship is $A \subset B$.