If and , then is equal to:

Q Explanation: To solve this, let's analyze the elements of sets and by substituting values of (where ). 1. Analyzing Set The general term for set is . For : For : For : For : So, 2. Analyzing Set The general term for set is , which can be written as . For : For : For : For : For : ... For : So, 3. Comparison and Conclusion From the values calculated: Set consists of all non-negative multiples of . Set consists of some specific multiples of (using binomial expansion, , so is always divisible by ). Every element in is also found in , but contains many elements (like ) that are not in . Therefore, (P is a subset of Q). When one set is a subset of another, their union is the larger set: Final Answer: The correct option is (c) .

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NIMCET 2009 — Mathematics PYQ

NIMCET | Mathematics | 2009

If $P = \{ (4^n - 3n - 1) \mid n \in N \}$ and $Q = \{ (9n - 9) \mid n \in N \}$ , then $P \cup Q$ is equal to:

Choose the correct answer:

Explanation

To solve this, let's analyze the elements of sets $P$ and $Q$ by substituting values of $n$ (where $n = 1, 2, 3, \dots$ ).

1. Analyzing Set $P$

The general term for set $P$ is $4^n - 3n - 1$ .

For $n = 1$ : $4^1 - 3(1) - 1 = 4 - 3 - 1 = 0$

For $n = 2$ : $4^2 - 3(2) - 1 = 16 - 6 - 1 = 9$

For $n = 3$ : $4^3 - 3(3) - 1 = 64 - 9 - 1 = 54$

For $n = 4$ : $4^4 - 3(4) - 1 = 256 - 12 - 1 = 243$

So, $P = \{0, 9, 54, 243, \dots\}$

2. Analyzing Set $Q$

The general term for set $Q$ is $9n - 9$ , which can be written as $9(n - 1)$ .

For $n = 1$ : $9(1 - 1) = 0$

For $n = 2$ : $9(2 - 1) = 9$

For $n = 3$ : $9(3 - 1) = 18$

For $n = 4$ : $9(4 - 1) = 27$

For $n = 5$ : $9(5 - 1) = 36$

...

For $n = 7$ : $9(7 - 1) = 54$

So, $Q = \{0, 9, 18, 27, 36, 45, 54, \dots\}$

3. Comparison and Conclusion

From the values calculated:

Set $Q$ consists of all non-negative multiples of $9$ .

Set $P$ consists of some specific multiples of $9$ (using binomial expansion, $4^n = (1+3)^n = 1 + 3n + \frac{n(n-1)}{2}3^2 + \dots$ , so $4^n - 3n - 1$ is always divisible by $9$ ).

Every element in $P$ is also found in $Q$ , but $Q$ contains many elements (like $18, 27, 36$ ) that are not in $P$ .

Therefore, $P \subset Q$ (P is a subset of Q).

When one set is a subset of another, their union is the larger set:

$P \cup Q = Q$

Final Answer:

The correct option is (c) $Q$ .