if A=`{5^(n)-4n-1: n in N}` and `B={16(n-1):n in N}` then

Options

A. A=B
B. `A cup B=phi`
C. `A subset B`
D. `B subset A`

Answer

To solve the problem, we need to analyze the sets \( A \) and \( B \) defined as follows:

- \( A = \{ 5^n - 4n - 1 : n \in \mathbb{N} \} \)
- \( B = \{ 16(n-1) : n \in \mathbb{N} \} \)

### Step 1: Calculate elements of set \( A \)

Let's compute the first few elements of set \( A \) by substituting natural numbers for \( n \).

1. For \( n = 1 \):
\[ A(1) = 5^1 - 4(1) - 1 = 5 - 4 - 1 = 0 \]

2. For \( n = 2 \):
\[ A(2) = 5^2 - 4(2) - 1 = 25 - 8 - 1 = 16 \]

3. For \( n = 3 \):
\[ A(3) = 5^3 - 4(3) - 1 = 125 - 12 - 1 = 112 \]

4. For \( n = 4 \):
\[ A(4) = 5^4 - 4(4) - 1 = 625 - 16 - 1 = 608 \]

Thus, the first few elements of set \( A \) are:
\[ A = \{ 0, 16, 112, 608, \ldots \} \]

### Step 2: Calculate elements of set \( B \)

Now, let's compute the first few elements of set \( B \):

1. For \( n = 1 \):
\[ B(1) = 16(1-1) = 16(0) = 0 \]

2. For \( n = 2 \):
\[ B(2) = 16(2-1) = 16(1) = 16 \]

3. For \( n = 3 \):
\[ B(3) = 16(3-1) = 16(2) = 32 \]

4. For \( n = 4 \):
\[ B(4) = 16(4-1) = 16(3) = 48 \]

Thus, the first few elements of set \( B \) are:
\[ B = \{ 0, 16, 32, 48, \ldots \} \]

### Step 3: Analyze the relationship between sets \( A \) and \( B \)

Now we can compare the two sets:

- The elements of \( A \) are \( 0, 16, 112, 608, \ldots \)
- The elements of \( B \) are \( 0, 16, 32, 48, \ldots \)

From the computed elements, we can observe that:
- \( 0 \) is in both sets.
- \( 16 \) is in both sets.
- \( 32 \) and \( 48 \) are in set \( B \) but not in set \( A \).
- \( 112 \) and \( 608 \) are in set \( A \) but not in set \( B \).

### Conclusion

From the above analysis, we can conclude that:
- \( A \cap B = \{ 0, 16 \} \)
- \( A \) and \( B \) are not equal.
- \( B \) is not a subset of \( A \) since it contains elements (like \( 32 \) and \( 48 \)) that are not in \( A \).

### Final Result

Thus, the correct relationship is that \( A \) and \( B \) have some common elements, but neither is a subset of the other.

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